Question

A tangent having a slope –$\frac{4}{3}$ to the ellipse $\frac{x^{2}}{18} + \frac{y^{2}}{32}$= 1 intersects the major and minor axes at A and B. If O is the origin, then the area DOAB is-

• A. 48 sq. units
• B. 9 sq. units
• C. 24 sq. units
• D. 16 sq. units

Answer 1

Answer: (C)

24 sq. units

Answer 2

Any point on the ellipse $\frac{x^{2}}{(3\sqrt{2})^{2}} + \frac{y^{2}}{(4\sqrt{2})^{2}}$ = 1 can be taken as $(3\sqrt{2}\cos{}\theta,4\sqrt{2}\sin\theta)$ and the slope of the tangent

= – $\frac{b^{2}x}{a^{2}y}$

= – $\frac{32(3\sqrt{2}\cos\theta)}{18(4\sqrt{2}\sin\theta)}$ = – $\frac{4}{3}$ cot q .... (1)

Given slope of the tangent = –$\frac{4}{3}$ .... (2)

From equations (1) and (2)

cot q = 1 Ž q = $\frac{\pi}{4}$

Hence the equation of the tangent is $\frac{x.\frac{1}{\sqrt{2}}}{3\sqrt{2}} + \frac{y.\frac{1}{\sqrt{2}}}{4\sqrt{2}}$= 1

(i.e.) $\frac{x}{6} + \frac{y}{8}$= 1

Hence A = (6, 0), B = (0, 8)

Area of DOAB = $\frac{1}{2}$× 6 × 8 = 24 sq. units