Question

The equation of the ellipse whose axes are coincident with the co-ordinate axes and which touches the straight lines

3x – 2y – 20 = 0 and x + 6y – 20 = 0, is –

• A. $\frac{x^{2}}{5} + \frac{y^{2}}{8}$ = 1
• B. $\frac{x^{2}}{40} + \frac{y^{2}}{10}$ = 10
• C. $\frac{x^{2}}{40} + \frac{y^{2}}{10}$ = 1
• D. $\frac{x^{2}}{10} + \frac{y^{2}}{40}$ = 1

Answer 1

Answer: (C)

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

Answer 2

Let the equation of the ellipse be

$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}}$ = 1.

We know that the general equation of the tangent to the ellipse is

y = mx ± $\sqrt{a^{2}m^{2} + b^{2}}$ … (1)

Since 3x – 2y – 20 = 0 i.e., 2y = 3x – 20 i.e.,
y = $\frac{3}{2}$x – 10, is tangent to the ellipse, therefore comparing with (1),

m = $\frac{3}{2}$ and a²m² + b² = 100

Ž a² · $\frac{9}{4}$ + b² = 100 Ž 9a² + 4b² = 400 … (2)

Similarly since x + 6y – 20 = 0 i.e., y = – $\frac{1}{6}$x + $\frac{10}{3}$

is tangent to the ellipse, therefore comparing with (1).

m = – $\frac{1}{6}$ and a²m² + b² = $\frac{100}{9}$

Ž $\frac{a^{2}}{36}$ + b² = $\frac{100}{9}$ Ž a² + 3b² = 400 … (3)

Now solving (2) and (3), we get a² = 40 and b² = 10.

\ The required equation of the ellipse is

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

Hence (3) is the correct answer.