The points of intersection of the perpendicul | vedclass

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(A) The given equation of the ellipse is $4x^2 + 9y^2 = 36$. Dividing by $36$,we get $\frac{x^2}{9} + \frac{y^2}{4} = 1$. Here,$a^2 = 9$ and $b^2 = 4$

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$x^2 + y^2 = 13$

Vedclass · Answer

(A) The given equation of the ellipse is $4x^2 + 9y^2 = 36$. Dividing by $36$,we get $\frac{x^2}{9} + \frac{y^2}{4} = 1$. Here,$a^2 = 9$ and $b^2 = 4$. The locus of the points of intersection of perpendicular tangents to an ellipse is known as its director circle. The equation of the director circle for an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is given by $x^2 + y^2 = a^2 + b^2$. Substituting the values of $a^2$ and $b^2$,we get $x^2 + y^2 = 9 + 4$. Therefore,the equation of the curve is $x^2 + y^2 = 13$.

The points of intersection of the perpendicular tangents drawn to the ellipse $4x^2 + 9y^2 = 36$ lie on the curve:

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