A tangent having slope of -frac{4}{3} to ellipse frac{{{x^2}}}{{18}} + frac{{{y^2}}}{{32}} = 1

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10-2. Parabola, Ellipse, Hyperbola

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A tangent having slope of $-\frac{4}{3}$ to the ellipse $\frac{{{x^2}}}{{18}}$ + $\frac{{{y^2}}}{{32}}$ $= 1$ intersects the major and minor axes in points $A$ and $ B$ respectively. If $C$ is the centre of the ellipse then the area of the triangle $ ABC$ is : .............. $\mathrm{sq. \,units}$

A

$12$

B

$24 $

C

$36$

D

$48 $

Solution

since the major axis is along the $y$ -axis.

$\therefore$ Equation of tangent is $x=m y+\sqrt{b^{2} m^{2}+a^{2}}$

Slope of tangent $=\frac{1}{m}=\frac{-4}{3} \Rightarrow m=\frac{-3}{4}$

Hence, equation of tangent is $4 x+3 y=24$ or $\frac{x}{6}+\frac{y}{8}=1$

Its intercepts on the axes are 6 and 8 .

Area $(\Delta A O B)=\frac{1}{2} \times 6 \times 8=24$ sq. unit

Standard 11

Mathematics

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