Question
The product of the perpendiculars drawn from the foci on any tangent to the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1,(a>b)$ is equal to(a) $a^{2}$(b) ab(c) $\mathrm{a}^{2} \mathrm{~b}^{2}$(d) $b^{2}$
Step 1
Step 1: The general equation of a tangent to the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ is given by $y=mx\pm\sqrt{a^{2}m^{2}+b^{2}}$. Show more…
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Tangent lines for an ellipse Show that an equation of the line tangent to the ellipse $x^{2} / a^{2}+y^{2} / b^{2}=1$ at the point $\left(x_{0}, y_{0}\right)$ is $$\frac{x x_{0}}{a^{2}}+\frac{y y_{0}}{b^{2}}=1$$.
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