ELLIPSE-5
Equations of Normals in different forms
(i) Point form
Let equation of ellipse be $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$
Equation of normal is $\frac{a^{2} x}{x _{1}}-\frac{b^{2} y}{y _{1}}=a^{2}-b^{2}$
(ii) Parametric form
Let equation of ellipse be $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$
Equation of normal is $a ~x\sec \phi-b ~ y\cosec \phi=\mathrm{a}^{2}-\mathrm{b}^{2}$
$(a \cos \phi, b \sin \phi)$ is parametric coordinates of $\mathrm{P}\left(\mathrm{x} _{1}, \mathrm{y} _{1}\right)$, i.e. $\mathrm{P}(\mathrm{a} \cos \phi, \mathrm{b} \sin \phi)$
(iii) Slope form
For an ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$
Equation of normal in terms of slope $(\mathrm{m})$ is
$ y=m x \pm \frac{m\left(a^{2}-b^{2}\right)}{\sqrt{a^{2}+b^{2} m^{2}}} $
Condition of normality when $y=m x+c$ is the normal of $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ is
$ \begin{array}{r} \mathrm{c}^{2}=\frac{\mathrm{m}^{2}\left(\mathrm{a}^{2}-\mathrm{b}^{2}\right)^{2}}{\mathrm{a}^{2}+\mathrm{b}^{2} \mathrm{~m}^{2}} \\ \text { or } \quad \mathrm{c}= \pm \frac{\mathrm{m}\left(\mathrm{a}^{2}-\mathrm{b}^{2}\right)}{\sqrt{\mathrm{a}^{2}+\mathrm{b}^{2} \mathrm{~m}^{2}}} \end{array} $
Examples
1. If the normal at the point $P(\theta)$ to the ellipse $\frac{x^{2}}{14}+\frac{y}{5}=1$ intersects it again at the point $Q(2 \theta)$ then $\cos \theta$ is equal to
(a) $\frac{2}{3}$
(b) $-\frac{2}{3}$
(c) $\frac{3}{2}$
(d) $-\frac{3}{2}$
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Solution :
The normal at $(a \sin \theta, b \sin \theta)$ is $\frac{a x}{\cos \theta}-\frac{b y}{\sin \theta}=a^{2}-b^{2}$
Here $\mathrm{a}^{2}=14 \& \mathrm{~b}^{2}=5$
$ a=\sqrt{14} \quad b=\sqrt{5} $
$\frac{\sqrt{14} \mathrm{x}}{\cos \theta}-\frac{\sqrt{5} \mathrm{y}}{\sin \theta}=14-5$
$\frac{\sqrt{14} \mathrm{x}}{\cos \theta}-\frac{\sqrt{5} \mathrm{y}}{\sin \theta}=9……..(1)$
$\therefore$ It meets the curve again at $\theta(2 \theta)(\operatorname{acos} 2 \theta, b \sin 2 \theta)$
i.e. $(\sqrt{14} \cos 2 \theta, \sqrt{5} \sin 2 \theta)$
$\therefore$ equation (1) satisfy this point
$\frac{14 \cos 2 \theta}{\cos \theta}-\frac{5 \sin 2 \theta}{\sin \theta}=9$
$\frac{14\left(2 \cos ^{2} \theta-1\right)}{\cos \theta}-\frac{5(2 \sin \theta \cos \theta)}{\sin \theta}=9$
$28 \cos ^{2} \theta-14-10 \cos ^{2} \theta=9 \cos \theta$
$18 \cos ^{2} \theta-9 \cos \theta-14=0$
$(6 \cos \theta-7)(3 \cos \theta+2)=0$
$\cos \theta=\frac{7}{6}>1$ not possible $(\because \cos \theta<1)$
$\cos \theta=\frac{-2}{3}$
$\therefore$ option (b) is correct.
2. The equation of the normal to the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ at the end of the latus rectum in the first quadrant is
(a) $x+$ ey $-a e^{3}=0$
(b) $ \mathrm{x}-\mathrm{ey}+\mathrm{ae} \mathrm{e}^{3}=0$
(c) $\mathrm{x}-\mathrm{ey}-\mathrm{ae}^{3}=0$
(d) None of these
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Solution :
The end of the latus rectum in the first quadrantal is $\left(\mathrm{ae}, \frac{\mathrm{b}^{2}}{\mathrm{a}}\right)$
Equation of normal at $\left(\mathrm{ae}, \frac{\mathrm{b}^{2}}{\mathrm{a}}\right)$ is
$ \begin{gathered} \frac{a^{2} x}{a e}-\frac{b^{2} y}{b^{2}}=a^{2}-b^{2}, \boxed{\frac{a^{2} x}{x _{1}}-\frac{b^{2} y}{y _{1}}=a^{2}-b^{2}} \\ \frac{a x}{e}-a y=a^{2}-b^{2} \\ a x-e a y=e a^{2}-e b^{2} \\ =e a^{2}-e\left(a^{2}-a^{2} e^{2}\right) \\ =e a^{2}-e a^{2}+a^{2} e^{3} \\ \therefore a x-a e y-a^{2} e^{3}=0 \\ x-e y-a e^{3}=0 \end{gathered} $
Correct option is c
3. The condition that the line $x \cos \alpha+y \sin \alpha=p$ may be a normal to the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ then
(a) $\left(a^{2}-b^{2}\right)^{2}=p^{2}\left(a^{2} \sec ^{2} \alpha+b^{2} \operatorname{cosec}^{2} \alpha\right)$
(c) $\left(a^{2}+b^{2}\right)^{2}=p^{2}\left(a^{2} \sec ^{2} \alpha+b^{2} \operatorname{cosec}^{2} \alpha\right)$
(b) $\left(\mathrm{a}^{2}-\mathrm{b}^{2}\right)^{2}=\mathrm{p}^{2}\left(\mathrm{a}^{2} \operatorname{cosec}^{2} \alpha+\mathrm{b}^{2} \sec ^{2} \alpha\right)$
(d) None of these
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Solution :
The equation of normal to $\frac{\mathrm{x}^{2}}{\mathrm{a}^{2}}+\frac{\mathrm{y}^{2}}{\mathrm{~b}^{2}}=1$ at ’ $\theta$ ’ is
$\frac{a^{2} x}{a \cos \theta}-\frac{b^{2} y}{b \sin \theta}-\left(a^{2}-b^{2}\right)=0 …….(1)$
Given that $x \cos \alpha+y \sin \alpha-p=0$ is normal to the ellipse comparing (1) & (2) we get
$ \begin{aligned} & \frac{\mathrm{a} / \cos \theta}{\cos \alpha}=\frac{-b / \sin \theta}{\sin \alpha}=\frac{a^{2}-b^{2}}{p} \\ & \text { or } \frac{a}{\cos \theta \cos \alpha}=\frac{-b}{\sin \theta \sin \alpha}=\frac{a^{2}-b^{2}}{p} \\ & \text { or } \frac{\cos \theta \cos \alpha}{a}=-\frac{\sin \theta \sin \alpha}{b}=\frac{p}{a^{2}-b^{2}} \\ & \cos \theta=\frac{a p}{\left(a^{2}-b^{2}\right) \cos \alpha}, \sin \theta=\frac{-b p}{\left(a^{2}-b^{2}\right) \sin \alpha} \end{aligned} $
Squaring and adding we get
$\cos ^{2} \theta+\sin ^{2} \theta=\frac{a^{2} p^{2}}{\left(a^{2}-b^{2}\right)^{2} \cos ^{2} \alpha}+\frac{b^{2} p^{2}}{\left(a^{2}-b^{2}\right)^{2} \sin ^{2} \alpha}$
$\left(\mathrm{a}^{2}-\mathrm{b}^{2}\right)^{2}=\mathrm{p}^{2}\left(\mathrm{a}^{2} \sec ^{2} \alpha+\mathrm{b}^{2} \operatorname{cosc}^{2} \alpha\right)$
$\therefore$ correct option is a
4. If the normals at $\mathrm{P}\left(\mathrm{x} _{1}, \mathrm{y} _{1}\right), \mathrm{Q}\left(\mathrm{x} _{2}, \mathrm{y} _{2}\right)$ and $\mathrm{R}\left(\mathrm{x} _{3}, \mathrm{y} _{3}\right)$ to the ellipse are concurrent, then
(a) $\left|\begin{array}{lll}x _{2} & y _{2} & x _{2} y _{2} \\ x _{1} & y _{1} & x _{1} y _{1} \\ x _{3} & y _{3} & x _{3} y _{3}\end{array}\right|=-1$
(b) $\left|\begin{array}{lll}\mathrm{x} _{1} & \mathrm{y} _{1} & \mathrm{x} _{1} \mathrm{y} _{1} \\ \mathrm{x} _{2} & \mathrm{y} _{2} & \mathrm{x} _{2} \mathrm{y} _{2} \\ \mathrm{x} _{3} & \mathrm{y} _{3} & \mathrm{x} _{3} \mathrm{y} _{3}\end{array}\right|=0$
(c) $\left|\begin{array}{lll}x _{2} & y _{2} & x _{2} y _{2} \\ x _{3} & y _{3} & x _{3} y _{3} \\ x _{1} & y _{1} & x _{1} y _{1}\end{array}\right|=1$
(d) None of these
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Solution :
The equations of the normals at $\mathrm{P}\left(\mathrm{x} _{1}, \mathrm{y} _{1}\right), \mathrm{Q}\left(\mathrm{x} _{2}, \mathrm{y} _{2}\right)$ and $\mathrm{R}\left(\mathrm{x} _{3}, \mathrm{y} _{3}\right)$ to the ellipse $\frac{\mathrm{x}^{2}}{\mathrm{a}^{2}}+\frac{\mathrm{y}^{2}}{\mathrm{~b}^{2}}=1$ are
$\frac{a^2 x}{x_1}-\frac{b^2 y}{y_1}=a^2-b^2………..(1)$
$\frac{a^2 x}{x_2}-\frac{b^2 y}{y_2}=a^2-b^2………..(2)$
$\frac{a^2 x}{x_3}-\frac{b^2 y}{y_3}=a^2-b^2………..(3)$
respectively
These lines are concurrent, if
$ \left|\begin{array}{lll} \frac{a^{2}}{x _{1}} & \frac{-b^{2}}{y _{1}} & a^{2}-b^{2} \\ \frac{a^{2}}{x _{2}} & \frac{-b^{2}}{y _{2}} & a^{2}-b^{2} \\ \frac{a^{2}}{x _{3}} & \frac{-b^{2}}{y _{3}} & a^{2}-b^{2} \end{array}\right|=0 $
$a^{2} b^{2}\left(a^{2}-b^{2}\right)\left|\begin{array}{ccc}\frac{1}{x _{1}} & \frac{-1}{y _{1}} & 1 \\ \frac{1}{x _{2}} & \frac{-1}{y _{2}} & 1 \\ \frac{1}{x _{3}} & \frac{-1}{y _{3}} & 1\end{array}\right|=0$
$\mathrm{R} _{1} \rightarrow \mathrm{x} _{1} \mathrm{y} _{1} \mathrm{R} _{1} ; \mathrm{R} _{2} \rightarrow \mathrm{x} _{2} \mathrm{y} _{2} \mathrm{R} _{2} ; \mathrm{R} _{3} \rightarrow \mathrm{x} _{3} \mathrm{y} _{3}$ we get
$ \left|\begin{array}{lll} \mathrm{y} _{1} & -\mathrm{x} _{1} & \mathrm{x} _{1} \mathrm{y} _{1} \\ \mathrm{y} _{2} & -\mathrm{x} _{2} & \mathrm{x} _{2} \mathrm{y} _{2} \\ \mathrm{y} _{3} & -\mathrm{x} _{3} & \mathrm{x} _{3} \mathrm{y} _{3} \end{array}\right|=0 $
OR $\left|\begin{array}{lll}x _{1} & y _{1} & x _{1} y _{1} \\ x _{2} & y _{2} & x _{2} y _{2} \\ x _{3} & y _{3} & x _{3} y _{3}\end{array}\right|=0$
your correct option is b
Practice questions
1. In the normal at the end of latus rectum of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ with eccentricity e, passes through one end of the minor axis, then :
(a) $\mathrm{e}^{2}\left(1+\mathrm{e}^{2}\right)=0$
(b) $\mathrm{e}^{2}\left(1+\mathrm{e}^{2}\right)=1$
(c) $\mathrm{e}^{2}\left(1+\mathrm{e}^{2}\right)=-1$
(d) $\mathrm{e}^{2}\left(1+\mathrm{e}^{2}\right)=2$
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Answer: (b)2. If the normals to $\frac{\mathrm{x}^{2}}{\mathrm{a}^{2}}+\frac{\mathrm{y}^{2}}{\mathrm{~b}^{2}}=1$ at the ends of the chords $\ell _{1} \mathrm{x}+\mathrm{m} _{1} \mathrm{y}=1$ and $\ell _{2} \mathrm{x}+\mathrm{m} _{2} \mathrm{y}=1$ are concurrent, then :
(a) $\mathrm{a}^{2} \ell _{1} \ell _{2}+\mathrm{b}^{2} \mathrm{~m} _{1} \mathrm{~m} _{2}=1$
(b) $\mathrm{a}^{2} \ell _{1} \ell _{2}+\mathrm{b} _{2} \mathrm{~m} _{1} \mathrm{~m} _{2}=-1$
(c) $\mathrm{a}^{2} \ell _{1} \ell _{2}-\mathrm{b}^{2} \mathrm{~m} _{1} \mathrm{~m} _{2}=-1$
(d) None of these
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Answer: (b)3. If the normal at an end of a latus rectum of an ellipse passes through one extremity of the minor axis, then the eccentricity of the ellipse is given by
(a) $\mathrm{e}^{4}+\mathrm{e}^{2}-1=0$
(b) $\mathrm{e}^{4}+\mathrm{e}^{2}-5=0$
(c) $\mathrm{e}^{3}=\sqrt{5}$
(d) None of these
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Answer: (a)4. The number of normals that can be drawn from a point to a given ellipse is
(a) 2
(b) 3
(c) 4
(d) 1
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Answer: (c)5. If the normal at any point $\mathrm{P}$ on the ellipse $\frac{\mathrm{x}^{2}}{\mathrm{a}^{2}}+\frac{\mathrm{y}^{2}}{\mathrm{~b}^{2}}=1$ meets the axes in $\mathrm{G}$ and $\mathrm{g}$ respectively, then $\mathrm{PG}: \mathrm{Pg}$ is equal to
(a) $a: b$
(b) $a^{2}: b^{2}$
(c) $b^{2}: a^{2}$
(d) $\mathrm{b}: \mathrm{a}$
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Answer: (c)6. If normal to ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ at $\left(a e, \frac{b^{2}}{a}\right)$ is passing through $(0,-2 b)$, then value of eccentricity is
(a) $\sqrt{2}-1$
(b) $2(\sqrt{2}-1)$
(c) $\sqrt{2(\sqrt{2}-1)}$
(d) None of these
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Answer: (c)7. If normal at any point $\mathrm{P}$ to the ellipse $\frac{\mathrm{x}^{2}}{\mathrm{a}^{2}}+\frac{\mathrm{y}^{2}}{\mathrm{~b}^{2}}=1, \mathrm{a}>\mathrm{b}$ meet the axes at $\mathrm{M}$ and $\mathrm{N}$ so that $\frac{\mathrm{PM}}{\mathrm{PN}}=\frac{2}{3}$, then value of eccentricity e is
(a) $\frac{1}{\sqrt{2}}$
(b) $\sqrt{\frac{2}{3}}$
(c) $\frac{1}{\sqrt{3}}$
(d) None of these
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Answer: (c)8. If the tangent drown at point $\left(t^{2}, 2 t\right)$ on the parabola $y^{2}=4 \mathrm{x}$ is same as the normal drawn at point $(\sqrt{5} \cos \theta, 2 \sin \theta)$ on the ellipse $4 x^{2}+5 y^{2}=20$. Then the values of $\operatorname{tand} \theta$ are
(a) $\theta=\cos ^{-1}\left(\frac{-1}{\sqrt{5}}\right) \& \mathrm{t}=\frac{-1}{\sqrt{5}}$
(b) $\theta=\cos ^{-1}\left(\frac{1}{\sqrt{5}}\right) \& \mathrm{t}=\frac{1}{\sqrt{5}}$
(c) $\theta=\cos ^{-1}\left(\frac{-2}{\sqrt{5}}\right) \& \mathrm{t}=\frac{-2}{\sqrt{5}}$
(d) None of these
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Answer: (a)9. The normals at four points on the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ meet in the point $(h, k)$. Then the mean position of the four points is
(a) $\left(\frac{a^{2} h}{2\left(a^{2}+b^{2}\right)}, \frac{b^{2} k}{2\left(a^{2}+b^{2}\right)}\right)$
(b) $\left(\frac{a^{3} h}{2\left(a^{2}+b^{2}\right)}, \frac{b^{3} k}{2\left(a^{2}+b^{2}\right)}\right)$
(c) $\left(\frac{\mathrm{ah}}{2\left(\mathrm{a}^{2}-\mathrm{b}^{2}\right)}, \frac{\mathrm{bk}}{2\left(\mathrm{a}^{2}-\mathrm{b}^{2}\right)}\right)$
(d) $\left(\frac{a^{2} h}{2\left(a^{2}-b^{2}\right)}, \frac{b^{2} k}{2\left(a^{2}-b^{2}\right)}\right)$
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Answer: (d)10. The equation of the normal at the point $(2,3)$ on the ellipse $9 x^{2}+16 y^{2}=180$ is
(a) $3 y=8 x-10$
(b) $3 y-8 x+7=0$
(c) $8 y+3 x+7=0$
(d) $3 x+2 y+7=0$
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Answer: (b)11. Number of distinct normal lines that can be drawn to the ellipse $\frac{x^{2}}{169}+\frac{y^{2}}{25}=1$ from the point $P$ $(0,6)$ is
(a) one
(b) two
(c) three
(d) four
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Answer: (c)12. Any ordinate MP of the ellipse $\frac{\mathrm{x}^{2}}{25}+\frac{\mathrm{y}^{2}}{9}=1$ meets the auxiliary circle at $\mathrm{Q}$, then locus of the point of intersection of normals at $\mathrm{P}$ and $\mathrm{Q}$ to the respective curve is
(a) $\mathrm{x}^{2}+\mathrm{y}^{2}=8$
(b) $x^{2}+y^{2}=34$
(c) $\mathrm{x}^{2}+\mathrm{y}^{2}=64$
(d) $x^{2}+y^{2}=15$
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Answer: (c)13. If the normals at $\mathrm{P}(\theta)$ and $\mathrm{Q}\left(\frac{\pi}{2}+\theta\right)$ to the ellipse $\frac{\mathrm{x}^{2}}{\mathrm{a}^{2}}+\frac{\mathrm{y}^{2}}{\mathrm{~b}^{2}}=1$ meet the major axis at $\mathrm{G}$ and $\mathrm{g}$ respectively, then $\mathrm{PG}^{2}+\mathrm{Qg}^{2}=$
(a) $\mathrm{b}^{2}\left(1-\mathrm{e}^{2}\right)\left(2-\mathrm{e}^{2}\right)$
(b) $\mathrm{a}^{2}\left(\mathrm{e}^{4}-\mathrm{e}^{2}+2\right)$
(c) $\mathrm{a}^{2}\left(1+\mathrm{e}^{2}\right)\left(2+\mathrm{e}^{2}\right)$
(d) $\mathrm{b}^{2}\left(1+\mathrm{e}^{2}\right)\left(2+\mathrm{e}^{2}\right)$
BETA
