Find the number of points on the ellipse frac | 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 c

Vedclass · Accepted Answer

$4$

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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 $\frac{x^{2}}{16} + \frac{y^{2}}{9} = 1$,we have $a^{2} = 16$ and $b^{2} = 9$.Thus,the equation of 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 $y = \pm \sqrt{\frac{50}{3}}$.Since there are two values for $x$ and for each $x$ there are two values for $y$,there are $4$ points of intersection.

Find the number of points on the ellipse $\frac{x^{2}}{50} + \frac{y^{2}}{20} = 1$ from which a pair of perpendicular tangents can be drawn to the ellipse $\frac{x^{2}}{16} + \frac{y^{2}}{9} = 1$.

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