Ellipse – JEE Advanced Previous Year Questions with Solutions

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Q. The line passing through the extremity A of the major axis and extremity B of the minor axis of the ellipse $x^{2}+9 y^{2}=9$ meets its auxiliary circle at the point M. Then the area of the triangle with vertices at A, M and the origin O is

(A) $\frac{31}{10}$

(B) $\frac{29}{10}$

(C) $\frac{21}{10}$

(D) $\frac{27}{10}$

[JEE 2009, 3]

Sol. (D)

Area of triangle $\mathrm{AOM}=\frac{1}{2} \mathrm{AO} . \mathrm{PM}$

$\Rightarrow$ Eqaution of $\mathrm{AM}$ is $\mathrm{y}=\frac{1}{3}(\mathrm{x}+3)$

$\mathrm{x}-3 \mathrm{y}+3=0$ which is chord of auxiliary circle $\mathrm{x}^{2}+\mathrm{y}^{2}=9$

and $\mathrm{PM}$ is ordinate of point $\mathrm{M}$

$\Rightarrow(3 \mathrm{y}-3)^{2}+\mathrm{y}^{2}=9 \Rightarrow \mathrm{y}=\frac{9}{5}=\mathrm{PM}$

$2 \cos \left(\frac{\mathrm{B}+\mathrm{C}}{2}\right) \cos \left(\frac{\mathrm{B}-\mathrm{C}}{2}\right)=4 \sin ^{2} \frac{\mathrm{A}}{2} \Rightarrow \cos \left(\frac{\mathrm{B}-\mathrm{C}}{2}\right)=2 \sin \frac{\mathrm{A}}{2}$

$\Rightarrow 2 \cos \frac{\mathrm{A}}{2} \cos \left(\frac{\mathrm{B}-\mathrm{C}}{2}\right)=2 \cdot 2 \sin \frac{\mathrm{A}}{2} \cdot \cos \frac{\mathrm{A}}{2} \Rightarrow 2 \sin \left(\frac{\mathrm{B}+\mathrm{C}}{2}\right) \cos \left(\frac{\mathrm{B}-\mathrm{C}}{2}\right)=2 \sin \mathrm{A}$

$\Rightarrow \sin \mathrm{B}+\sin \mathrm{C}=2 \sin \mathrm{A}$

$\Rightarrow \mathrm{b}+\mathrm{c}=2 \mathrm{a}$

$\Rightarrow$ locus is ellipse

Q. The normal at a point $P$ on the ellipse $x^{2}+4 y^{2}=16$ meets the $x$ -axis at $Q .$ If $M$ is the mid point of the line segment $P Q$, then the locus of Mintersects the latus rectums of the given ellipse at the point

(A) $\left(\pm \frac{3 \sqrt{5}}{2}, \pm \frac{2}{7}\right)$

(B) $\left(\pm \frac{3 \sqrt{5}}{2}, \pm \frac{\sqrt{19}}{4}\right)$

(C) $\left(\pm 2 \sqrt{3}, \pm \frac{1}{7}\right)$

(D) $\left(\pm 2 \sqrt{3}, \pm \frac{4 \sqrt{3}}{7}\right)$

[JEE 2009, 3]

Sol. (C)

Q. In a triangle PQR, let $\angle \mathrm{PQR}=30^{\circ}$ and the sides $\mathrm{PQ}$ and $\mathrm{QR}$ have lengths $10 \sqrt{3}$ and 10 , respectively. Then, which of the following statement(s) is (are) TRUE?

(A) $\angle \mathrm{QPR}=45^{\circ}$

(B) The area of the triangle PQR is $25 \sqrt{3}$ and $\angle \mathrm{QRP}=120^{\circ}$

(C) The radius of the incircle of the triangle PQR is $10 \sqrt{3}-15$

(D) The area of the circumcircle of the triangle PQR is $100 \pi$.

Sol. (B,C,D)

$2 \cos \left(\frac{\mathrm{B}+\mathrm{C}}{2}\right) \cos \left(\frac{\mathrm{B}-\mathrm{C}}{2}\right)=4 \sin ^{2} \frac{\mathrm{A}}{2} \Rightarrow \cos \left(\frac{\mathrm{B}-\mathrm{C}}{2}\right)=2 \sin \frac{\mathrm{A}}{2}$

$\Rightarrow 2 \cos \frac{\mathrm{A}}{2} \cos \left(\frac{\mathrm{B}-\mathrm{C}}{2}\right)=2 \cdot 2 \sin \frac{\mathrm{A}}{2} \cdot \cos \frac{\mathrm{A}}{2} \Rightarrow 2 \sin \left(\frac{\mathrm{B}+\mathrm{C}}{2}\right) \cos \left(\frac{\mathrm{B}-\mathrm{C}}{2}\right)=2 \sin \mathrm{A}$

$\Rightarrow \sin \mathrm{B}+\sin \mathrm{C}=2 \sin \mathrm{A}$

$\Rightarrow \mathrm{b}+\mathrm{c}=2 \mathrm{a}$

$\Rightarrow$ locus is ellipse.

Comprehension: $4 \mathrm{to} 5$

Tangents are drawn from the point $\mathrm{P}(3,4)$ to the ellipse $\frac{\mathrm{x}^{2}}{9}+\frac{\mathrm{y}^{2}}{4}=1$ touching the ellipse at points A and B

Q. The coordinates of A and B are

(A) (3,0) and (0,2)

(B) $\left(-\frac{8}{5}, \frac{2 \sqrt{261}}{15}\right)$ and $\left(-\frac{9}{5}, \frac{8}{5}\right)$

(C) $\left(-\frac{8}{5}, \frac{2 \sqrt{161}}{15}\right)$ and $(0,2)$

(D) $(3,0)$ and $\left(-\frac{9}{5}, \frac{8}{5}\right)$

[JEE 2010, 3+3+3]

Sol. (D)

Q. The orthocenter of the triangle PAB is

(A) $\left(5, \frac{8}{7}\right)$

(B) $\left(\frac{7}{5}, \frac{25}{8}\right)$

(C) $\left(\frac{11}{5}, \frac{8}{5}\right)$

(D) $\left(\frac{8}{25}, \frac{7}{5}\right)$

Sol. (C)

Q. The equation of the locus of the point whose distances from the point P and the line AB are equal, is –

(A) $9 x^{2}+y^{2}-6 x y-54 x-62 y+241=0$

(B) $x^{2}+9 y^{2}+6 x y-54 x+62 y-241=0$

(C) $9 x^{2}+9 y^{2}-6 x y-54 x-62 y-241=0$

(D) $x^{2}+y^{2}-2 x y+27 x+31 y-120=0$

Sol. (A)

Equation of line $\mathrm{AB}$ is $\mathrm{x}+3 \mathrm{y}=3$

Now let the point be $(\mathrm{h}, \mathrm{k})$

According to question,

$\left|\frac{h+3 k-3}{\sqrt{1^{2}+3^{2}}}\right|=\sqrt{(h-3)^{2}+(4-k)^{2}}$

After solving, we get

$9 x^{2}+y^{2}-6 x y-54 x-62 y+241=0$

Q. The ellipse $\mathrm{E}_{1}: \frac{\mathrm{x}^{2}}{9}+\frac{\mathrm{x}^{2}}{4}=1$ is inscribed in a rectangle $\mathrm{R}$ whose sides are parallel to the coordinate axes. Another ellipse $\mathrm{E}_{2}$ passing through the point $(0,4)$ circumscribes the rectangle $\mathrm{R}$. The eccentricity of the ellipse $\mathrm{E}_{2}$ is $-$

(A) $\frac{\sqrt{2}}{2}$

(B) $\frac{\sqrt{3}}{2}$

(C) $\frac{1}{2}$

(D) $\frac{3}{4}$

[JEE 2012, 3M, –1M]

Sol. (C)

Q. A vertical line passing through the point (h,0) intersects the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$ at the points $P$ and $Q$. Let the tangents to the ellipse at $P$ and $Q$ meet at the point $R .$ If $\Delta(h)=$ area of the triangle $P Q R, \Delta_{1}=\max _{1 / 2 \leq h \leq 1} \Delta(h)$ and $\Delta_{2}=\min _{1 / 2 \Delta \operatorname{Lh} \leq 1} \Delta(h),$ then $\frac{8}{\sqrt{5}} \Delta_{1}-8 \Delta_{2}=$

Sol. (9)

Q.

Sol. (A)

Q. Suppose that the foci of the ellipse $\frac{x^{2}}{9}+\frac{y^{2}}{5}=1$ are $\left(f_{1}, 0\right)$ and $\left(f_{2}, 0\right)$ where $f_{1}>0$ and $f_{2}<0 .$ Let $\mathrm{P}_{1}$ and $\mathrm{P}_{2}$ be two parabolas with a common vertex at $(0,0)$ and with foci at $\left(f_{1}, 0\right)$ and $\left(2 f_{2}, 0\right),$ respectively. Let $\mathrm{T}_{1}$ be a tangent to $\mathrm{P}_{1}$ which passes through $\left(2 f_{2}, 0\right)$ and $\mathrm{T}_{2}$ be a tangent to $\mathrm{P}_{2}$ which passes through $\left(f_{1}, 0\right) .$ If $\mathrm{m}_{1}$ is the slope of $\mathrm{T}_{1}$ and $\mathrm{m}_{2}$ is the slope of $\mathrm{T}_{2},$ then the value of

$\left(\frac{1}{\mathrm{m}_{1}^{2}}+\mathrm{m}_{2}^{2}\right)$ is

[JEE 2015, 4M, –0M]

Sol. 4

Q. Let $\mathrm{E}_{1}$ and $\mathrm{E}_{2}$ be two ellipses whose centers are at the origin. The major axes of $\mathrm{E}_{1}$ and $\mathrm{E}_{2}$ lie along the $\mathrm{x}$ -axis and the $\mathrm{y}$ -axis, respectively. Let $\mathrm{S}$ be the circle $\mathrm{x}^{2}+(\mathrm{y}-1)^{2}=2 .$ The straight line $\mathrm{x}+\mathrm{y}=3$ touches the curves $\mathrm{S}, \mathrm{E}_{1}$ and $\mathrm{E}_{2}$ at $\mathrm{P}, \mathrm{Q}$ and $\mathrm{R},$ respectively. Suppose that $\mathrm{PQ}=\mathrm{PR}=\frac{2 \sqrt{2}}{3} .$ If $\mathrm{e}_{1}$ and $\mathrm{e}_{2}$ are the eccentricities of $\mathrm{E}_{1}$ and $\mathrm{E}_{2},$ respectively, then the correct expression(s) is(are)

(A) $\mathrm{e}_{1}^{2}+\mathrm{e}_{2}^{2}=\frac{43}{40}$

(B) $\mathrm{e}_{1} \mathrm{e}_{2}=\frac{\sqrt{7}}{2 \sqrt{10}}$

(C) $\left|\mathrm{e}_{1}^{2}-\mathrm{e}_{2}^{2}\right|=\frac{5}{8}$

(D) $\mathrm{e}_{1} \mathrm{e}_{2}=\frac{\sqrt{3}}{4}$

[JEE 2015, 4M, –0M]

Sol. (A,B)

Let $\mathrm{E}_{1}: \frac{\mathrm{x}^{2}}{\mathrm{a}^{2}}+\frac{\mathrm{y}^{2}}{\mathrm{b}^{2}}=1 \quad(\mathrm{a}>\mathrm{b})$

$\mathrm{\&} \quad \mathrm{E}_{2}: \frac{\mathrm{x}^{2}}{\mathrm{c}^{2}}+\frac{\mathrm{y}^{2}}{\mathrm{d}^{2}}=1 \quad(\mathrm{c}<\mathrm{d})$

$\& \quad \mathrm{S}: \mathrm{x}^{2}+(\mathrm{y}-1)^{2}=2$

$\mathrm{\&} \quad$ tangent to $\mathrm{E}_{1}, \mathrm{E}_{2} \& \mathrm{S}$ is $\mathrm{x}+\mathrm{y}=3$

Now, point of contact of $\mathrm{S} \&$ tangent is $\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)$

Let $\mathrm{x}=\mathrm{X} \quad \& \mathrm{y}-1=\mathrm{Y}$

$\therefore \quad \mathrm{X}^{2}+\mathrm{Y}^{2}=2$

$\mathrm{\&} \quad \mathrm{X}+\mathrm{Y}=2$

Let $\left(\mathrm{X}_{1}, \mathrm{Y}_{1}\right)$ be point of contact.

$\therefore \quad \mathrm{X}_{1}+\mathrm{YY}_{1}=2$

$\therefore \quad \mathrm{X}_{1}=1 \& \mathrm{Y}_{1}=1$

$\therefore \quad \mathrm{x}_{1}=1 \& \mathrm{y}_{1}=2$

Now, parametric equation of $x+y=3$

is $\frac{x-1}{-\frac{1}{\sqrt{2}}}=\frac{y-2}{\frac{1}{\sqrt{2}}}=\pm \frac{2 \sqrt{2}}{3} \Rightarrow x=\frac{5}{3}, y=\frac{4}{3}$

$\Rightarrow \quad x=\frac{1}{3} \& y=\frac{8}{3}$

$\therefore \quad \mathrm{P} \equiv(1,2), \quad \mathrm{Q} \equiv\left(\frac{5}{3}, \frac{4}{3}\right) \quad \& \quad \mathrm{R}=\left(\frac{1}{3}, \frac{8}{3}\right)$

Now, equation tangent at $\mathrm{Q}$ on ellipse $\mathrm{E}_{1}$

$\frac{\mathrm{x} \cdot 5}{3 \mathrm{a}^{2}}+\frac{\mathrm{y} \cdot 4}{3 \mathrm{b}^{2}}=1$

Comparing it with $\mathrm{x}+\mathrm{y}=3$

$\therefore \quad \mathrm{a}^{2}=5 \& \mathrm{b}^{2}=4$

Now, $e_{1}^{2}=1-\frac{4}{5}=\frac{1}{5}$

Similarly, $e_{2}^{2}=\frac{7}{8}$

$\therefore \quad e_{1}^{2} e_{2}^{2}=\frac{7}{40} \Rightarrow e_{1} e_{2}=\frac{\sqrt{7}}{2 \sqrt{10}}$

$e_{1}^{2}+e_{2}^{2}=\frac{1}{5}+\frac{7}{8}=\frac{43}{40} ; \quad\left|e_{1}^{2}-e_{2}^{2}\right|=\left|\frac{1}{5}-\frac{7}{8}\right|=\frac{27}{40}$

$\therefore \quad(A) \&(B)$

Let $\mathrm{F}_{1}\left(\mathrm{x}_{1}, 0\right)$ and $\mathrm{F}_{2}\left(\mathrm{x}_{2}, 0\right)$ for $\mathrm{x}_{1}<0$ and $\mathrm{x}_{2}>0,$ be the foci of the ellipse $\frac{\mathrm{x}^{2}}{9}+\frac{\mathrm{y}^{2}}{8}=1 .$ Suppose a parabola having vertex at the origin and focus at $\mathrm{F}_{2}$ intersects the ellipse at point $\mathrm{M}$ in the first quadrant and at point $\mathrm{N}$ in the fourth quadrant.

Q. The orthocentre of the triangle $F_{1} \mathrm{MN}$ is-

(A) $\left(-\frac{9}{10}, 0\right)$

(B) $\left(\frac{2}{3}, 0\right)$

(C) $\left(\frac{9}{10}, 0\right)$

(D) $\left(\frac{2}{3}, \sqrt{6}\right)$

[JEE 2015, 4M, –0M]

Sol. (A)

Orthocentre lies on $\mathrm{x}$ -axis

Equation of altitude through $\mathrm{M}: \mathrm{y}-\sqrt{6}=\frac{5}{2 \sqrt{6}}\left(\mathrm{x}-\frac{3}{2}\right)$

Equation of altitude through $\mathrm{F}_{1}: \mathrm{y}=0$

solving, we get orthocentre $\left(-\frac{9}{10}, 0\right)$

Q. If the tangents to the ellipse at M and N meet at R and the normal to the parabola at M meets the x-axis at Q, then the ratio of area of the triangle MQR to area of the quadrilateral $\mathrm{MF}_{1} \mathrm{NF}_{2}$ is-

(A) 3 : 4 (B) 4 : 5 (C) 5 : 8 (D) 2 : 3

Sol. (C)

Normal to parabola at $\mathrm{M}: \mathrm{y}-\sqrt{6}=-\frac{\sqrt{6}}{2.1}\left(\mathrm{x}-\frac{3}{2}\right)$

Solving it with $\mathrm{y}=0,$ we get $\mathrm{Q} \equiv\left(\frac{7}{2}, 0\right)$

Tangent to ellipse at $M: \frac{x \cdot \frac{3}{2}}{9}+\frac{y(\sqrt{6})}{8}=1$

Solving it with $y=0$, we get $R \equiv(6,0)$

$\therefore$ Area of triangle MQR $=\frac{1}{2} \cdot\left(6-\frac{7}{2}\right) \cdot \sqrt{6}=\frac{5 \sqrt{6}}{4}$

Area of quadrilateral $\mathrm{MF}_{1} \mathrm{NF}_{2}=2 \cdot \frac{1}{2} \cdot(1-(-1)) \cdot \sqrt{6}=2 \sqrt{6}$

Required ratio $=5: 8$

Q. Consider two straight lines, each of which is tangent to both the circle $x^{2}+y^{2}=\frac{1}{2}$ and the parabola $\mathrm{y}^{2}=4 \mathrm{x}$. Let these lines intersect at the point Q. Consider the ellipse whose center is at the origin O(0, 0) and whose semi-major axis is OQ. If the length of the minor axis of this ellipse is $\sqrt{2}$, then the which of the following statement(s) is (are) TRUE ?

(A) For the ellipse, the eccentricity is $\frac{1}{\sqrt{2}}$ and the length of the latus rectum is 1

(B) For the ellipse, the eccentricity is $\frac{1}{2}$ and the length of the latus rectum is $\frac{1}{2}$

(C) The area of the region bounded by the ellipse between the lines $\mathrm{x}=\frac{1}{\sqrt{2}}$ and $\mathrm{x}=1$ is $\frac{1}{4 \sqrt{2}}(\pi-2)$

(D) The area of the region bounded by the ellipse between the lines $x=\frac{1}{\sqrt{2}}$ and $x=1$ is $\frac{1}{16}(\pi-2)$

Sol. (A,C)

Let equation of common tangent is $\mathrm{y}=\mathrm{mx}+\frac{1}{\mathrm{m}}$

$\therefore \quad\left|\frac{0+0+\frac{1}{\mathrm{m}}}{\sqrt{1+\mathrm{m}^{2}}}\right|=\frac{1}{\sqrt{2}} \Rightarrow \mathrm{m}^{4}+\mathrm{m}^{2}-2=0 \Rightarrow \mathrm{m}=\pm 1$

Equation of common tangents are $\mathrm{y}=\mathrm{x}+1$ and $\mathrm{y}=-\mathrm{x}-1$

point $\mathrm{Q}$ is $(-1,0)$

$\therefore$ Equation of ellipse is $\frac{\mathrm{x}^{2}}{1}+\frac{\mathrm{y}^{2}}{1 / 2}=1$

(A) $\mathrm{e}=\sqrt{1-\frac{1}{2}}=\frac{1}{\sqrt{2}}$ and $\mathrm{LR}=\frac{2 \mathrm{b}^{2}}{\mathrm{a}}=1$

Area $2 . \int_{1 / \sqrt{2}}^{1} \frac{1}{\sqrt{2}} \cdot \sqrt{1-x^{2}} d x=\sqrt{2}\left[\frac{x}{2} \sqrt{1-x^{2}}+\frac{1}{2} \sin ^{-1} x\right]_{1 / \sqrt{2}}^{1}$

$=\sqrt{2}\left[\frac{\pi}{4}-\left(\frac{1}{4}+\frac{\pi}{8}\right)\right]=\sqrt{2}\left(\frac{\pi}{8}-\frac{1}{4}\right)=\frac{\pi-2}{4 \sqrt{2}}$

correct answer are (A) and (D)