If a tangent having slope of - frac{4}{3} to | vedclass

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Question

(d) Let $P({x_1},{y_1})$ be a point on the ellipse

$\frac{{{x^2}}}{{18}} + \frac{{{y^2}}}{{32}} = 1$ ==>$\frac{{x_1^2}}{{18}} + \frac{{y_1^2}}{{

Vedclass · Accepted Answer

$24$

Vedclass · Answer

(d) Let $P({x_1},{y_1})$ be a point on the ellipse

$\frac{{{x^2}}}{{18}} + \frac{{{y^2}}}{{32}} = 1$ ==>$\frac{{x_1^2}}{{18}} + \frac{{y_1^2}}{{32}} = 1$

The equation of the tangent at $({x_1},{y_1})$ is $\frac{{x{x_1}}}{{18}} + \frac{{y{y_1}}}{{32}} = 1$.

This meets the axes at $A\left( {\frac{{18}}{{{x_1}}},\,0} ight)$ and $B\left( {0,\,\frac{{32}}{{{y_1}}}} ight)$. It is given

that slope of the tangent at $({x_1},{y_1})$ is $ - \frac{3}{4}$.

Hence $ - \frac{{{x_1}}}{{18}}.\frac{{32}}{{{y_1}}} = - \frac{4}{3}$ ==>$\frac{{{x_1}}}{{{y_1}}} = \frac{3}{4}$ ==>$\frac{{{x_1}}}{3} = \frac{{{y_1}}}{4} = k$ $(say)$

$	herefore {x_1} = 3k,\,\,{y_1} = 4k$

Putting ${x_1},{y_1}$in (i), we get ${k^2} = 1$.

Now area of $\Delta OAB = \frac{1}{2}OA.OB = \frac{1}{2}\frac{{18}}{{{x_1}}}.\frac{{32}}{{{y_1}}} = \frac{1}{2}\frac{{(18)(32)}}{{({x_1}{y_1})}}$

$ = \frac{1}{2}\frac{{(18)(32)}}{{(3k)(4k)}} = \frac{{24}}{{{k^2}}} = 24$ sq. unit , .

If 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, then the area of $\Delta OAB$ is equal to .................. $\mathrm{sq. \, units}$ ($O$ is centre of the ellipse)

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