Number of points on the ellipse frac{x^2}{50} | vedclass

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(D) The locus of points from which perpendicular tangents can be drawn to an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is its director circle,gi

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$4$

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(D) The locus of points from which perpendicular tangents can be drawn to an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is its director circle,given by $x^2 + y^2 = a^2 + b^2$.For the ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$,the director circle is $x^2 + y^2 = 16 + 9 = 25$.We need to find the number of intersection points of this director circle $x^2 + y^2 = 25$ and the given ellipse $\frac{x^2}{50} + \frac{y^2}{20} = 1$.From the circle,$y^2 = 25 - x^2$. Substituting this into the ellipse equation:$\frac{x^2}{50} + \frac{25 - x^2}{20} = 1$Multiplying by $100$:$2x^2 + 5(25 - x^2) = 100$$2x^2 + 125 - 5x^2 = 100$$-3x^2 = -25$$x^2 = \frac{25}{3}$,which gives $x = \pm \frac{5}{\sqrt{3}}$.Since $x^2 = \frac{25}{3}  0$,which gives $y = \pm \sqrt{\frac{50}{3}}$.Thus,there are $4$ distinct points of intersection.

Number of points on the ellipse $\frac{x^2}{50} + \frac{y^2}{20} = 1$ from which a pair of perpendicular tangents are drawn to the ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$ is:

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