The tangents drawn from the point P(3, 4) to | vedclass

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(A) The equation of the chord of contact $AB$ for the point $P(3, 4)$ with respect to the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$ is given by $T =

Vedclass · Accepted Answer

$9x^2 + y^2 - 6xy - 54x - 62y + 241 = 0$

Vedclass · Answer

(A) The equation of the chord of contact $AB$ for the point $P(3, 4)$ with respect to the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$ is given by $T = 0$.Substituting $(x_1, y_1) = (3, 4)$,we get $\frac{3x}{9} + \frac{4y}{4} = 1$,which simplifies to $\frac{x}{3} + y = 1$,or $x + 3y = 3$.Let the point be $(h, k)$. The condition that the distance from $(h, k)$ to $P(3, 4)$ is equal to the distance from $(h, k)$ to the line $x + 3y - 3 = 0$ is:$\sqrt{(h-3)^2 + (k-4)^2} = \frac{|h + 3k - 3|}{\sqrt{1^2 + 3^2}}$.Squaring both sides: $(h-3)^2 + (k-4)^2 = \frac{(h + 3k - 3)^2}{10}$.$10(h^2 - 6h + 9 + k^2 - 8k + 16) = h^2 + 9k^2 + 9 + 6hk - 6h - 18k$.$10h^2 + 10k^2 - 60h - 80k + 250 = h^2 + 9k^2 + 6hk - 6h - 18k + 9$.$9h^2 + k^2 - 6hk - 54h - 62k + 241 = 0$.Replacing $(h, k)$ with $(x, y)$,the locus is $9x^2 + y^2 - 6xy - 54x - 62y + 241 = 0$.

The tangents drawn from the point $P(3, 4)$ to the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$ touch the ellipse at points $A$ and $B$. The equation of the locus of a point which is equidistant from point $P$ and the line $AB$ is:

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