Q1. \(\dfrac{x^2}{36}+\dfrac{y^2}{16}=1\)

Solution

The given equation of the ellipse is \(\dfrac{x^{2}}{36}+\dfrac{y^{2}}{16}=1\). Since the larger denominator is under \(x^{2}\), the major axis lies along the \(x\)-axis. Comparing with the standard form \(\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=1\), we obtain \(a^{2}=36\) and \(b^{2}=16\).

\[ \begin{aligned} a^{2}&=36\\ a&=6\\ b^{2}&=16\\ b&=4\\ c&=\sqrt{a^{2}-b^{2}}\\ &=\sqrt{6^{2}-4^{2}}\\ &=\sqrt{36-16}\\ c&=\sqrt{20} \end{aligned} \]

The foci of an ellipse whose major axis lies along the \(x\)-axis are given by \((\pm c,0)\). Hence, the coordinates of the foci are \[\left(\pm\sqrt{20},0\right)\]

The vertices of the ellipse lie at the ends of the major axis. Therefore, the coordinates of the vertices are

\[ \begin{aligned} (\pm a,0)=(\pm 6,0) \end{aligned} \]

The length of the major axis of an ellipse is equal to \(2a\). Hence, its length is

\[ \begin{aligned} 2a&=2\times 6\\ &=12 \end{aligned} \]

The length of the minor axis of an ellipse is equal to \(2b\). Therefore, the length of the minor axis is

\[ \begin{aligned} 2b&=2\times 4\\ &=8 \end{aligned} \]

The eccentricity of an ellipse is defined as the ratio of the distance of a focus from the centre to the semi-major axis. Thus, the eccentricity is

\[ \begin{aligned} e&=\dfrac{c}{a}\\&=\dfrac{\sqrt{20}}{6}\\ &=\dfrac{\sqrt{5}}{3} \end{aligned} \]

The length of the latus rectum of an ellipse whose major axis lies along the \(x\)-axis is given by \(\dfrac{2b^{2}}{a}\). Hence, the length of the latus rectum is

\[ \begin{aligned} \dfrac{2b^{2}}{a}&=\dfrac{2\times 4^{2}}{6}\\ &=\dfrac{16}{3} \end{aligned} \]

Q2. \(\dfrac{x^2}{4}+\dfrac{y^2}{25}=1\)

Solution

Given ellipse:

$$ \begin{aligned} \dfrac{x^{2}}{4}+\dfrac{y^{2}}{25}=1 \end{aligned} $$ Since the larger denominator is under \(y^2\), the major axis lies along the \(y\)-axis

$$ \begin{aligned} a^{2}&=4 \\ \Rightarrow a&=2 \\ b^{2}&=25 \\ \Rightarrow b&=5 \end{aligned} $$ Here \(a\) is the semi-minor axis and \(b\) is the semi-major axis

Foci

$$ \begin{aligned} &\left(0,\pm c\right) \\ c&=\sqrt{b^{2}-a^{2}} \\ &=\sqrt{25-4} \\ &=\sqrt{21} \\ \Rightarrow &\left(0,\pm \sqrt{21}\right) \end{aligned} $$

Vertices

$$ \begin{aligned} \left(0,\pm b\right)\\ =\left(0,\pm 5\right) \end{aligned} $$

Length of Major Axis

$$ \begin{aligned} 2b &= 2\times 5 \\ &= 10 \end{aligned} $$

Length of Minor Axis

$$ \begin{aligned} 2a &= 2\times 2 \\ &= 4 \end{aligned} $$

Eccentricity

$$ \begin{aligned} e &= \dfrac{c}{b} \\ &= \dfrac{\sqrt{21}}{5} \end{aligned} $$

Length of Latus Rectum

$$ \begin{aligned} \dfrac{2a^{2}}{b} &= \dfrac{2\times 2^{2}}{5} \\ &= \dfrac{8}{5} \end{aligned} $$

Q3. \(\dfrac{x^2}{16}+\dfrac{y^2}{9}=1\)

Solution

Given ellipse:

$$ \begin{aligned} \dfrac{x^{2}}{16}+\dfrac{y^{2}}{9}=1 \end{aligned} $$ Since the larger denominator is under \(x^2\), the major axis lies along the \(x\)-axis

$$ \begin{aligned} a^{2}&=16 \\ \Rightarrow a&=4 \\ b^{2}&=9 \\ \Rightarrow b&=3 \end{aligned} $$ Here \(a\) is the semi-major axis and \(b\) is the semi-minor axis

Foci of Ellipse

$$ \begin{aligned} &\left( \pm c,0\right) \\ c&=\sqrt{a^{2}-b^{2}} \\ &=\sqrt{16-9} \\ &=\sqrt{7} \\ \Rightarrow &\left( \pm \sqrt{7},0\right) \end{aligned} $$

Vertices

$$ \begin{aligned} &\left( \pm a,0\right) \\ &=\left( \pm 4,0\right) \end{aligned} $$

Length of Major Axis

$$ \begin{aligned} 2a &= 2\times 4 \\ &= 8 \end{aligned} $$

Length of Minor Axis

$$ \begin{aligned} 2b &= 2\times 3 \\ &= 6 \end{aligned} $$

Eccentricity

$$ \begin{aligned} e &= \dfrac{c}{a} \\ &= \dfrac{\sqrt{7}}{4} \end{aligned} $$

Latus Rectum

$$ \begin{aligned} \dfrac{2b^{2}}{a} &= \dfrac{2\times 9}{4} \\ &= \dfrac{9}{2} \end{aligned} $$

Q4. \(\dfrac{x^2}{25}+\dfrac{y^2}{100}=1\)

Solution

Given ellipse:

$$ \begin{aligned} \dfrac{x^{2}}{25}+\dfrac{y^{2}}{100}=1 \end{aligned} $$ Since the larger denominator is under \(y^2\), the major axis lies along the \(y\)-axis

$$ \begin{aligned} a^{2}&=25 \\ \Rightarrow a&=5 \\ b^{2}&=100 \\ \Rightarrow b&=10 \end{aligned} $$ Here \(a\) represents the semi-minor axis and \(b\) represents the semi-major axis

Foci of Ellipse

$$ \begin{aligned} c&=\sqrt{b^{2}-a^{2}} \\ &=\sqrt{100-25} \\ &=\sqrt{75} \\ &=5\sqrt{3} \\ &\left( 0,\pm c\right) \\ \Rightarrow &\left( 0,\pm 5\sqrt{3}\right) \end{aligned} $$

Vertices

$$ \begin{aligned} &\left( 0,\pm b\right) \\ \Rightarrow &\left( 0,\pm 10\right) \end{aligned} $$

Length of Major Axis

$$ \begin{aligned} 2b &= 2\times 10 \\ &= 20 \end{aligned} $$

Length of Minor Axis

$$ \begin{aligned} 2a &= 2\times 5 \\ &= 10 \end{aligned} $$

Eccentricity

$$ \begin{aligned} e&=\dfrac{c}{b} \\ &=\dfrac{5\sqrt{3}}{10} \\ &=\dfrac{\sqrt{3}}{2} \end{aligned} $$

Latus Rectum

$$ \begin{aligned} \dfrac{2a^{2}}{b}&=\dfrac{2\times 25}{10} \\ &= 5 \end{aligned} $$

Q5. \(\dfrac{x^2}{49}+\dfrac{y^2}{36}=1\)

Solution

Given ellipse:

$$ \begin{aligned} \dfrac{x^{2}}{49}+\dfrac{y^{2}}{36}=1 \end{aligned} $$ Since the larger denominator is under \(x^2\), the major axis lies along the \(x\)-axis

$$ \begin{aligned} a^{2}&=49 \\ \Rightarrow a&=7 \\ b^{2}&=36 \\ \Rightarrow b&=6 \\ c&=\sqrt{a^{2}-b^{2}} \\ &=\sqrt{49-36} \\ &=\sqrt{13} \end{aligned} $$ Here \(a\) is the semi-major axis and \(b\) is the semi-minor axis

Foci of Ellipse

$$ \begin{aligned} &\left( \pm c,0\right) \\ \Rightarrow &\left( \pm \sqrt{13},0\right) \end{aligned} $$

Vertices

$$ \begin{aligned} &\left( \pm a,0\right) \\ &=\left( \pm 7,0\right) \end{aligned} $$

Length of Major Axis

$$ \begin{aligned} 2a &= 2\times 7 \\ &= 14 \end{aligned} $$

Length of Minor Axis

$$ \begin{aligned} 2b &= 2\times 6 \\ &= 12 \end{aligned} $$

Eccentricity

$$ \begin{aligned} e &= \dfrac{c}{a} \\ &= \dfrac{\sqrt{13}}{7} \end{aligned} $$

Latus Rectum

$$ \begin{aligned} \dfrac{2b^{2}}{a} &= \dfrac{2\times 36}{7} \\ &= \dfrac{72}{7} \end{aligned} $$

Q6. \(\dfrac{x^2}{100}+\dfrac{y^2}{400}=1\)

Solution

Given ellipse:

$$ \begin{aligned} \dfrac{x^{2}}{100}+\dfrac{y^{2}}{400}=1 \end{aligned} $$ Since the larger denominator is under \(y^2\), the major axis lies along the \(y\)-axis

$$ \begin{aligned} a^{2}&=100 \\ \Rightarrow a&=10 \\ b^{2}&=400 \\ \Rightarrow b&=20 \\ c&=\sqrt{b^{2}-a^{2}} \\ &=\sqrt{400-100} \\ &=\sqrt{300} \\ &=10\sqrt{3} \end{aligned} $$ Here \(a\) represents the semi-minor axis and \(b\) represents the semi-major axis

Foci of Ellipse

$$ \begin{aligned} &\left( 0,\pm c\right) \\ &=\left( 0,\pm 10\sqrt{3}\right) \end{aligned} $$

Vertices

$$ \begin{aligned} &\left( 0,\pm b\right) \\ &=\left( 0,\pm 20\right) \end{aligned} $$

Length of Major Axis

$$ \begin{aligned} 2b &= 2\times 20 \\ &= 40 \end{aligned} $$

Length of Minor Axis

$$ \begin{aligned} 2a &= 2\times 10 \\ &= 20 \end{aligned} $$

Eccentricity

$$ \begin{aligned} e&=\dfrac{c}{b} \\ &=\dfrac{10\sqrt{3}}{20} \\ &=\dfrac{\sqrt{3}}{2} \end{aligned} $$

Latus Rectum

$$ \begin{aligned} \dfrac{2a^{2}}{b}&=\dfrac{2\times 100}{20} \\ &= 10 \end{aligned} $$

Q7. \(36x^2 + 4y^2 = 144\)

Solution

Given ellipse:

$$ \begin{aligned} 36x^{2}+4y^{2}&=144 \\ \dfrac{36x^{2}}{144}+\dfrac{4y^{2}}{144}&=1 \\ \dfrac{x^{2}}{4}+\dfrac{y^{2}}{36}&=1 \end{aligned} $$ Since the larger denominator is under \(y^2\), the major axis lies along the \(y\)-axis

$$ \begin{aligned} a^{2}&=4 \\ \Rightarrow a&=2 \\ b^{2}&=36 \\ \Rightarrow b&=6 \\ c&=\sqrt{b^{2}-a^{2}} \\ &=\sqrt{36-4} \\ &=\sqrt{32} \\ &=4\sqrt{2} \end{aligned} $$ Here \(a\) represents the semi-minor axis and \(b\) represents the semi-major axis

Foci of Ellipse

$$ \begin{aligned} &\left( 0,\pm c\right) \\ &=\left( 0,\pm 4\sqrt{2}\right) \end{aligned} $$

Vertices

$$ \begin{aligned} &\left( 0,\pm b\right) \\ &=\left( 0,\pm 6\right) \end{aligned} $$

Length of Major Axis

$$ \begin{aligned} 2b &= 2\times 6 \\ &= 12 \end{aligned} $$

Length of Minor Axis

$$ \begin{aligned} 2a &= 2\times 2 \\ &= 4 \end{aligned} $$

Eccentricity

$$ \begin{aligned} e&=\dfrac{c}{b} \\ &=\dfrac{4\sqrt{2}}{6} \\ &=\dfrac{2\sqrt{2}}{3} \end{aligned} $$

Latus Rectum

$$ \begin{aligned} \dfrac{2a^{2}}{b}&=\dfrac{2\times 4}{6} \\ &=\dfrac{4}{3} \end{aligned} $$

Q8. \(16x^2 + y^2 = 16\)

Solution

Given ellipse:

$$ \begin{aligned} 16x^{2}+y^{2}&=16 \\ x^{2}+\dfrac{y^{2}}{16}&=1 \end{aligned} $$ Since the larger denominator is under \(y^2\), the major axis lies along the \(y\)-axis

$$ \begin{aligned} a^{2}&=1 \\ \Rightarrow a&=1 \\ b^{2}&=16 \\ \Rightarrow b&=4 \\ c&=\sqrt{b^{2}-a^{2}} \\ &=\sqrt{16-1} \\ &=\sqrt{15} \end{aligned} $$ Here \(a\) represents the semi-minor axis and \(b\) represents the semi-major axis

Foci of Ellipse

$$ \begin{aligned} &\left( 0,\pm c\right) \\ &=\left( 0,\pm \sqrt{15}\right) \end{aligned} $$

Vertices

$$ \begin{aligned} &\left( 0,\pm b\right) \\ &=\left( 0,\pm 4\right) \end{aligned} $$

Length of Major Axis

$$ \begin{aligned} 2b &= 2\times 4 \\ &= 8 \end{aligned} $$

Length of Minor Axis

$$ \begin{aligned} 2a &= 2\times 1 \\ &= 2 \end{aligned} $$

Eccentricity

$$ \begin{aligned} e&=\dfrac{c}{b} \\ &=\dfrac{\sqrt{15}}{4} \end{aligned} $$

Latus Rectum

$$ \begin{aligned} \dfrac{2a^{2}}{b}&=\dfrac{2\times 1}{4} \\ &=\dfrac{1}{2} \end{aligned} $$

Q9. \(4x^2 + 9y^2 = 36\)

Solution

Given ellipse:

$$ \begin{aligned} 4x^{2}+9y^{2}&=36 \\ \dfrac{4x^{2}}{36}+\dfrac{9y^{2}}{36}&=1 \\ \dfrac{x^{2}}{9}+\dfrac{y^{2}}{4}&=1 \end{aligned} $$ Since the larger denominator is under \(x^2\), the major axis lies along the \(x\)-axis

$$ \begin{aligned} a^{2}&=9 \\ \Rightarrow a&=3 \\ b^{2}&=4 \\ \Rightarrow b&=2 \\ c&=\sqrt{a^{2}-b^{2}} \\ &=\sqrt{9-4} \\ &=\sqrt{5} \end{aligned} $$ Here \(a\) represents the semi-major axis and \(b\) represents the semi-minor axis

Foci

$$ \begin{aligned} &\left( \pm c,0\right) \\ &=\left( \pm \sqrt{5},0\right) \end{aligned} $$

Vertices

$$ \begin{aligned} &\left( \pm a,0\right) \\ &=\left( \pm 3,0\right) \end{aligned} $$

Length of Major Axis

$$ \begin{aligned} 2a &= 2\times 3 \\ &= 6 \end{aligned} $$

Length of Minor Axis

$$ \begin{aligned} 2b &= 2\times 2 \\ &= 4 \end{aligned} $$

Eccentricity

$$ \begin{aligned} e&=\dfrac{c}{a} \\ &=\dfrac{\sqrt{5}}{3} \end{aligned} $$

Latus Rectum

$$ \begin{aligned} \dfrac{2b^{2}}{a}&=\dfrac{2\times 4}{3} \\ &=\dfrac{8}{3} \end{aligned} $$

Q10. Vertices (± 5, 0), foci (± 4, 0)

Solution

Given data:

$$ \begin{aligned} V&=\left( \pm 5,0\right) \\ F&=\left( \pm 4,0\right) \\ a&=5 \\ c&=4 \end{aligned} $$ Since the vertices and foci lie on the \(x\)-axis, the major axis is along the \(x\)-axis

$$ \begin{aligned} c^{2}&=a^{2}-b^{2} \\ 16&=25-b^{2} \\ b^{2}&=25-16 \\ b^{2}&=9 \\ \Rightarrow b&=3 \end{aligned} $$ Here \(a\) is the semi-major axis and \(b\) is the semi-minor axis

Equation of Ellipse

$$ \dfrac{x^{2}}{25}+\dfrac{y^{2}}{9}=1 $$

Q11. Vertices (0, ± 13), foci (0, ± 5)

Solution

Given data:

$$ \begin{aligned} V&=\left( 0,\pm 13\right) \\ F&=\left( 0,\pm 5\right) \\ b&=13 \\ c&=5 \end{aligned} $$ Since the vertices and foci lie on the \(y\)-axis, the major axis is along the \(y\)-axis

$$ \begin{aligned} c^{2}&=b^{2}-a^{2} \\ a^{2}&=b^{2}-c^{2} \\ &=13^{2}-5^{2} \\ &=169-25 \\ &=144 \\ \Rightarrow a&=12 \end{aligned} $$ Here \(b\) is the semi-major axis and \(a\) is the semi-minor axis

Equation of Ellipse

$$ \begin{aligned} \dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}&=1 \\ \dfrac{x^{2}}{144}+\dfrac{y^{2}}{169}&=1 \end{aligned} $$

Q12. Vertices (± 6, 0), foci (± 4, 0)

Solution

Given data:

$$ \begin{aligned} V&=\left( \pm 6,0\right) \\ F&=\left( \pm 4,0\right) \\ a&=6 \\ c&=4 \end{aligned} $$ Since the vertices and foci lie on the \(x\)-axis, the major axis is along the \(x\)-axis

$$ \begin{aligned} c^{2}&=a^{2}-b^{2} \\ \Rightarrow b^{2}&=a^{2}-c^{2} \\ b^{2}&=36-16 \\ &=20 \end{aligned} $$ Here \(a\) represents the semi-major axis and \(b\) represents the semi-minor axis

Equation of Ellipse

$$ \begin{aligned} \dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}&=1 \\ \dfrac{x^{2}}{36}+\dfrac{y^{2}}{20}&=1 \end{aligned} $$

Q13. Ends of major axis (± 3, 0), ends of minor axis (0, ± 2)

Solution

Given data:

Ends of the major axis are \(\left( \pm 3,0\right)\) and ends of the minor axis are \(\left( 0,\pm 2\right)\) Since the major axis lies along the \(x\)-axis, the ellipse is centered at the origin with semi-major axis along \(x\)

$$ \begin{aligned} a&=3 \\ b&=2 \end{aligned} $$ Here \(a\) represents the semi-major axis and \(b\) represents the semi-minor axis

Equation of Ellipse

$$ \begin{aligned} \dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}&=1 \\ \dfrac{x^{2}}{9}+\dfrac{y^{2}}{4}&=1 \end{aligned} $$

Q14. Ends of major axis \((0, ±\sqrt{5})\), ends of minor axis \((± 1, 0)\)

Solution

Given data:

End of the major axis $$ \begin{aligned} &\left( 0,\pm \sqrt{5}\right) \\ \Rightarrow b&=\sqrt{5} \\ \Rightarrow b^{2}&=5 \end{aligned} $$ Since the major axis lies along the \(y\)-axis, \(b\) represents the semi-major axis

Ends of the minor axis $$ \begin{aligned} &\left( \pm 1,0\right) \\ \Rightarrow a&=1 \\ \Rightarrow a^{2}&=1 \end{aligned} $$ Here \(a\) represents the semi-minor axis

Equation of Ellipse

$$ \begin{aligned} \dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}&=1 \\ \dfrac{x^{2}}{1}+\dfrac{y^{2}}{5}&=1 \end{aligned} $$

Q15. Length of major axis 26, foci (± 5, 0)

Solution

Given data:

Length of the major Axis = 26 $$ \begin{aligned} a&=\dfrac{26}{2} \\ a&=13 \\ a^{2}&=169 \end{aligned} $$ Since the length of the major axis is twice the semi-major axis, \(a\) represents the semi-major axis

Foci of ellipse

$$ \begin{aligned} &\left( \pm 5,0\right) \\ c&=5 \\ c^{2}&=a^{2}-b^{2} \\ 5^{2}&=13^{2}-b^{2} \\ b^{2}&=13^{2}-5^{2} \\ &=169-25 \\ &=144 \end{aligned} $$ Here \(c\) is the distance of each focus from the center and \(b\) is the semi-minor axis

Equation of Ellipse

$$ \begin{aligned} \dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}&=1 \\ \dfrac{x^{2}}{169}+\dfrac{y^{2}}{144}&=1 \end{aligned} $$

Q16. Length of minor axis 16, foci (0, ± 6).

Solution

Given data:

Length of minor axis = 16 $$ \begin{aligned} F&=\left( 0,\pm 6\right) \\ a&=\dfrac{16}{2} \\ &=8 \\ c&=6 \\ c^{2}&=b^{2}-a^{2} \\ 36&=b^{2}-64 \\ b^{2}&=36+64 \\ &=100 \end{aligned} $$ Since the foci lie on the \(y\)-axis, the major axis is along the \(y\)-axis, where \(b\) is the semi-major axis and \(a\) is the semi-minor axis

Equation of Ellipse

$$ \begin{aligned} \dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}&=1 \\ \dfrac{x^{2}}{64}+\dfrac{y^{2}}{100}&=1 \end{aligned} $$

Q17. Foci (± 3, 0), a = 4

Solution

Given data:

$$ \begin{aligned} F&=\left( \pm 3,0\right) \\ a&=4 \\ c&=3 \end{aligned} $$ Since the foci lie on the \(x\)-axis, the major axis is along the \(x\)-axis and \(a\) represents the semi-major axis

$$ \begin{aligned} c^{2}&=a^{2}-b^{2} \\ 9&=16-b^{2} \\ b^{2}&=16-9 \\ &=7 \end{aligned} $$ Here \(b\) represents the semi-minor axis

Equation of Ellipse

$$ \begin{aligned} \dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}&=1 \\ \dfrac{x^{2}}{16}+\dfrac{y^{2}}{7}&=1 \end{aligned} $$

Q18. \(b = 3,\; c = 4\), centre at the origin; foci on the x axis.

Solution

Given data:

\(b = 3,\; c = 4\), centre at the origin and foci on the \(x\)-axis, so the major axis lies along the \(x\)-axis and \(a\) represents the semi-major axis

$$ \begin{aligned} c^{2}&=a^{2}-b^{2} \\ 4^{2}&=a^{2}-3^{2} \\ a^{2}&=16+9 \\ &=25 \end{aligned} $$

Equation of Ellipse

$$ \begin{aligned} \dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}&=1 \\ \dfrac{x^{2}}{25}+\dfrac{y^{2}}{9}&=1 \end{aligned} $$

Q19. Centre at (0,0), major axis on the y-axis and passes through the points (3, 2) and (1,6).

Solution

Given data:

Centre \(=\,(0,0)\), major axis along the \(y\)-axis, so the standard form of the ellipse is

$$ \begin{aligned} \dfrac{x^{2}}{b^{2}}+\dfrac{y^{2}}{a^{2}}=1 \end{aligned} $$

Since the ellipse passes through \((3,2)\) and \((1,6)\), substitute these points into the equation

$$ \begin{aligned} \dfrac{3^{2}}{b^{2}}+\dfrac{2^{2}}{a^{2}}&=1 \\ \dfrac{9}{b^{2}}+\dfrac{4}{a^{2}}&=1 \end{aligned} $$

$$ \begin{aligned} \dfrac{1^{2}}{b^{2}}+\dfrac{6^{2}}{a^{2}}&=1 \\ \dfrac{1}{b^{2}}+\dfrac{36}{a^{2}}&=1 \end{aligned} $$

Let \(x=\dfrac{1}{b^{2}}\) and \(y=\dfrac{1}{a^{2}}\)

$$ \begin{aligned} 9x+4y&=1 \\ x+36y&=1 \end{aligned} $$

Multiply the second equation by \(9\)

$$ \begin{aligned} 9x+324y=9 \end{aligned} $$

Subtract the first equation

$$ \begin{aligned} -320y&=-8 \\ y&=\dfrac{8}{320} \\ &=\dfrac{1}{40} \end{aligned} $$

Substitute \(y=\dfrac{1}{40}\) into \(9x+4y=1\)

$$ \begin{aligned} 9x+4\left(\dfrac{1}{40}\right)&=1 \\ 9x+\dfrac{1}{10}&=1 \\ 9x&=\dfrac{9}{10} \\ x&=\dfrac{1}{10} \end{aligned} $$

$$ \begin{aligned} \dfrac{1}{b^{2}}&=\dfrac{1}{10} \\\Rightarrow b^{2}&=10 \\ \dfrac{1}{a^{2}}&=\dfrac{1}{40} \\\Rightarrow a^{2}&=40 \end{aligned} $$

Equation of the ellipse

$$ \dfrac{x^{2}}{10}+\dfrac{y^{2}}{40}=1 $$

Q20. Major axis on the x-axis and passes through the points (4,3) and (6,2).

Solution

Given data:

Major axis lies on the \(x\)-axis, so the standard form of the ellipse is

$$ \begin{aligned} \dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=1 \end{aligned} $$

Since the ellipse passes through \((4,3)\) and \((6,2)\), substitute these points

$$ \begin{aligned} \dfrac{4^{2}}{a^{2}}+\dfrac{3^{2}}{b^{2}}&=1 \\ \dfrac{16}{a^{2}}+\dfrac{9}{b^{2}}&=1 \\ \dfrac{6^{2}}{a^{2}}+\dfrac{2^{2}}{b^{2}}&=1 \\ \dfrac{36}{a^{2}}+\dfrac{4}{b^{2}}&=1 \end{aligned} $$

Let \(x=\dfrac{1}{a^{2}}\) and \(y=\dfrac{1}{b^{2}}\)

$$ \begin{aligned} 16x+9y&=1 \\ 36x+4y&=1 \end{aligned} $$

Multiply the first equation by \(4\) and the second by \(9\)

$$ \begin{aligned} 64x+36y&=4 \\ 324x+36y&=9 \end{aligned} $$

Subtract the equations to eliminate \(y\)

$$ \begin{aligned} -260x&=-5 \\ x&=\dfrac{5}{260} \\ x&=\dfrac{1}{52} \end{aligned} $$

Substitute \(x=\dfrac{1}{52}\) into \(16x+9y=1\)

$$ \begin{aligned} 16\left(\dfrac{1}{52}\right)+9y&=1 \\ \dfrac{16}{52}+9y&=1 \\ 9y&=1-\dfrac{16}{52} \\ 9y&=\dfrac{36}{52} \\ y&=\dfrac{4}{52} \\ y&=\dfrac{1}{13} \end{aligned} $$

$$ \begin{aligned} \dfrac{1}{a^{2}}=\dfrac{1}{52} \Rightarrow a^{2}&=52 \\ \dfrac{1}{b^{2}}=\dfrac{1}{13} \Rightarrow b^{2}&=13 \end{aligned} $$

Equation of ellipse

$$ \begin{aligned} \dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}&=1 \\ \dfrac{x^{2}}{52}+\dfrac{y^{2}}{13}&=1 \end{aligned} $$