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  • The tangents drawn from a point P to the ellipse <img alt="\mathrm{ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%20%5Cfrac%7Bx%5E2%7D%7

The tangents drawn from a point P to the ellipse \mathrm{ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1} make angles \mathrm{ \theta_1 ~and ~\theta_2}  with the major axis; the locus of P when \mathrm{ \tan ^2 \theta_1+\tan ^2 \theta_2=\Lambda} (a constant) is \mathrm{ k\left(x^2 y^2+a^2 y^2+x^2 b^2-a^2 b^2\right)=\lambda\left(x^2-a^2\right)^2}, where \mathrm{ k}=

Option: 1

\frac{1}{2}


Option: 2

2


Option: 3

4


Option: 4

\frac{1}{4}


Answers (1)

best_answer

Any tangent to the ellipse \mathrm{ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 is y=m x \pm \sqrt{a^2 m^2+b^2}}

\begin{aligned} &\mathrm{ \text { Let } P \equiv(h, k)=\quad k=m h \pm \sqrt{a^2 m^2+b^2} \Rightarrow \quad(k-m h)^2=a^2 m^2+b^2 }\\ &\mathrm{ \Rightarrow \quad m^2\left(h^2-a^2\right)-2 h k m+\left(k^2-b^2\right)=0 \quad \quad \quad \dots(i)}\\ &\mathrm{ \Rightarrow \quad m_1+m_2=\frac{2 h k}{h^2-a^2} \text { and } m_1 m_2=\frac{k^2-b^2}{h^2-a^2}\quad \quad \quad \dots(ii)} \end{aligned}

If \theta_1 and \theta_2 are the angles of inclination of tangents to the x-axis, then

\mathrm{\tan \theta_1+\tan \theta_2=\frac{2 h k}{h^2-a^2} \text { and } \tan \theta_1 \tan \theta_2=\frac{k^2-b^2}{h^2-a^2}\quad \quad \quad \quad \dots(iii)}

Given that \mathrm{ \tan ^2 \theta_1+\tan ^2 \theta_2=\lambda \Rightarrow\left(\tan \theta_1+\tan \theta_2\right)^2-2 \tan \theta_5 \tan \theta_2=\lambda }
\begin{aligned} &\mathrm{ =\quad\left(\frac{2 h k}{h^2-a^2}\right)^2-\frac{2\left(k^2-b^2\right)}{h^2-a^2}=\lambda }\\ &\mathrm{ =\quad 4 h^2 k^2-2\left(k^2-b^2\right)\left(h^2-a^2\right)=\lambda\left(h^2-a^2\right)_2 }\\ &\mathrm{ =\quad 2\left(h^2 k^2+k^2 a^2+h^2 b^2-a^2 b^2\right)=\lambda\left(h^2-a^2\right)^2 }\\ &\mathrm{ =\quad \text { Required locus is } 2\left(x^2 y^2+a^2 y^2+x^2 b^2-a^2 b^2\right)=\lambda\left(x^2-a^2\right)^2} \end{aligned}

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sudhir.kumar

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