If 4x - 3y - 5 = 0 is a normal to the ellipse | vedclass

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(B) The equation of the ellipse is $3x^2 + 8y^2 = k$,which can be written as $\frac{x^2}{k/3} + \frac{y^2}{k/8} = 1$.Let the point of contact be $(x_1

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$3x - 4y + 10 = 0$

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(B) The equation of the ellipse is $3x^2 + 8y^2 = k$,which can be written as $\frac{x^2}{k/3} + \frac{y^2}{k/8} = 1$.Let the point of contact be $(x_1, y_1)$. The equation of the normal at $(x_1, y_1)$ is $\frac{a^2x}{x_1} - \frac{b^2y}{y_1} = a^2 - b^2$.Here $a^2 = k/3$ and $b^2 = k/8$,so $\frac{kx}{3x_1} - \frac{ky}{8y_1} = \frac{k}{3} - \frac{k}{8} = \frac{5k}{24}$.Dividing by $k$,we get $\frac{x}{3x_1} - \frac{y}{8y_1} = \frac{5}{24}$,or $\frac{8x}{x_1} - \frac{3y}{y_1} = 5$.Comparing this with $4x - 3y - 5 = 0$,we get $\frac{8}{x_1} = 4 \Rightarrow x_1 = 2$ and $\frac{3}{y_1} = 3 \Rightarrow y_1 = 1$.Since $(2, 1)$ lies on the ellipse,$3(2)^2 + 8(1)^2 = k \Rightarrow k = 12 + 8 = 20$.For the point $(-2, m)$ on the ellipse,$3(-2)^2 + 8m^2 = 20$ $\Rightarrow 12 + 8m^2 = 20$ $\Rightarrow 8m^2 = 8$ $\Rightarrow m = 1$ (as $m > 0$).The tangent at $(-2, 1)$ to $3x^2 + 8y^2 = 20$ is given by $3x(-2) + 8y(1) = 20$.$-6x + 8y = 20 \Rightarrow 3x - 4y + 10 = 0$.

If $4x - 3y - 5 = 0$ is a normal to the ellipse $3x^2 + 8y^2 = k$,then the equation of the tangent drawn to this ellipse at the point $(-2, m)$ $(m > 0)$ is

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