iCON Education HYD, 79930 92826, 73309 72826JEE Main 2021 (Online) 26th August Morning Shift
On the ellipse ${{{x^2}} \over 8} + {{{y^2}} \over 4} = 1$ let P be a point in the second quadrant such that the tangent at P to the ellipse is perpendicular to the line x + 2y = 0. Let S and S' be the foci of the ellipse and e be its eccentricity. If A is the area of the triangle SPS' then, the value of (5 $-$ e2). A is :
A.
6
B.
12
C.
14
D.
24
Correct Answer: A
Explanation:
Equation of tangent : y = 2x + 6 at P
$\therefore$ P($-$8/3, 2/3)
$e = {1 \over {\sqrt 2 }}$
S & S' = ($-$2, 0) & (2, 0)
Area of $\Delta$SPS' = ${1 \over 2} \times 4 \times {2 \over 3}$
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2021 (Online) 27th July Morning Shift
A ray of light through (2, 1) is reflected at a point P on the y-axis and then passes through the point (5, 3). If this reflected ray is the directrix of an ellipse with eccentricity ${1 \over 3}$ and the distance of the nearer focus from this directrix is ${8 \over {\sqrt {53} }}$, then the equation of the other directrix can be :
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2021 (Online) 25th July Evening Shift
If a tangent to the ellipse x2 + 4y2 = 4 meets the tangents at the extremities of it major axis at B and C, then the circle with BC as diameter passes through the point :
A.
$(\sqrt 3 ,0)$
B.
$(\sqrt 2 ,0)$
C.
(1, 1)
D.
($-$1, 1)
Correct Answer: A
Explanation:
${{{x^2}} \over 4} + {{{y^2}} \over 1} = 1$
Equation of tangent i (cos$\theta$)x + 2sin$\theta$y = 2
so, $\left( {\sqrt 3 ,0} \right)$ satisfying option (1)
2021
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2021 (Online) 25th July Morning Shift
Let an ellipse $E:{{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$, ${a^2} > {b^2}$, passes through $\left( {\sqrt {{3 \over 2}} ,1} \right)$ and has eccentricity ${1 \over {\sqrt 3 }}$. If a circle, centered at focus F($\alpha$, 0), $\alpha$ > 0, of E and radius ${2 \over {\sqrt 3 }}$, intersects E at two points P and Q, then PQ2 is equal to :
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2021 (Online) 22th July Evening Shift
Let ${E_1}:{{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1,a > b$. Let E2 be another ellipse such that it touches the end points of major axis of E1 and the foci of E2 are the end points of minor axis of E1. If E1 and E2 have same eccentricities, then its value is :
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2021 (Online) 18th March Evening Shift
Let a tangent be drawn to the ellipse ${{{x^2}} \over {27}} + {y^2} = 1$ at $(3\sqrt 3 \cos \theta ,\sin \theta )$ where $0 \in \left( {0,{\pi \over 2}} \right)$. Then the value of $\theta$ such that the sum of intercepts on axes made by this tangent is minimum is equal to :
also f'($\theta$) changes sign $-$ to + hence minimum.
2021
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2021 (Online) 16th March Evening Shift
If the points of intersections of the ellipse ${{{x^2}} \over {16}} + {{{y^2}} \over {{b^2}}} = 1$ and the circle x2 + y2 = 4b, b > 4 lie on the curve y2 = 3x2, then b is equal to :
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2020 (Online) 6th September Evening Slot
If the normal at an end of a latus rectum of an
ellipse passes through an extremity of the
minor axis, then the eccentricity e of the ellipse
satisfies :
A.
e4 + 2e2 – 1 = 0
B.
e4 + e2 – 1 = 0
C.
e2 + 2e – 1 = 0
D.
e2 + e – 1 = 0
Correct Answer: B
Explanation:
Equation of normal at $\left( {ae,{{{b^2}} \over a}} \right)$
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2020 (Online) 6th September Morning Slot
Which of the following points lies on the locus of the foot of perpedicular drawn upon any tangent
to the ellipse,
${{{x^2}} \over 4} + {{{y^2}} \over 2} = 1$
from any of its foci?
A.
$\left( { - 1,\sqrt 3 } \right)$
B.
$\left( { - 2,\sqrt 3 } \right)$
C.
$\left( { - 1,\sqrt 2 } \right)$
D.
$\left( {1,2 } \right)$
Correct Answer: A
Explanation:
Let foot of perpendicular is (h, k)
Given ${{{x^2}} \over 4} + {{{y^2}} \over 2} = 1$
$ \therefore $ a = 2, b = $\sqrt 2 $
and e = $\sqrt {1 - {2 \over 4}} $ = ${1 \over {\sqrt 2 }}$
$ \therefore $ $\left( { - 1,\sqrt 3 } \right)$ lies on the locus.
2020
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2020 (Online) 5th September Morning Slot
If the co-ordinates of two points A and B are $\left( {\sqrt 7 ,0} \right)$ and $\left( { - \sqrt 7 ,0} \right)$ respectively and P is any
point on the conic, 9x2 + 16y2 = 144, then PA + PB is equal to :
Now e = $\sqrt {1 - {9 \over {16}}} = {{\sqrt 7 } \over 4}$
A and B are foci
PA + PB = 2a = 2 × 4 = 8
2020
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2020 (Online) 4th September Evening Slot
Let x = 4 be a directrix to an ellipse whose centre is at the origin and its eccentricity is ${1 \over 2}$. If P(1, $\beta $), $\beta $ > 0 is a point on this ellipse, then the equation of the normal to it at P is :
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2020 (Online) 4th September Morning Slot
Let ${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$ (a > b) be a given ellipse, length of whose latus rectum is 10. If its eccentricity is the maximum value of the function, $\phi \left( t \right) = {5 \over {12}} + t - {t^2}$, then a2 + b2 is equal to :
A.
145
B.
126
C.
135
D.
116
Correct Answer: B
Explanation:
Given ellipse ${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$ (a > b)
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2020 (Online) 9th January Evening Slot
The length of the minor axis (along y-axis) of
an ellipse in the standard form is ${4 \over {\sqrt 3 }}$. If this
ellipse touches the line, x + 6y = 8; then its
eccentricity is :
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2020 (Online) 8th January Morning Slot
Let the line y = mx and the ellipse 2x2 + y2 = 1
intersect at a ponit P in the first quadrant. If the
normal to this ellipse at P meets the co-ordinate axes at $\left( { - {1 \over {3\sqrt 2 }},0} \right)$ and (0, $\beta $), then $\beta $ is equal
to :
A.
${{\sqrt 2 } \over 3}$
B.
${2 \over 3}$
C.
${{2\sqrt 2 } \over 3}$
D.
${2 \over {\sqrt 3 }}$
Correct Answer: A
Explanation:
Let P be (x1
, y1)
Equation of normal at P is ${x \over {2{x_1}}} - {y \over {{y_1}}} = {1 \over 2} - 1$
It passes through $\left( { - {1 \over {3\sqrt 2 }},0} \right)$
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2020 (Online) 7th January Evening Slot
If 3x + 4y = 12$\sqrt 2 $ is a tangent to the ellipse
${{{x^2}} \over {{a^2}}} + {{{y^2}} \over 9} = 1$ for some $a$ $ \in $ R, then the distance
between the foci of the ellipse is :
A.
$2\sqrt 5 $
B.
$2\sqrt 7 $
C.
4
D.
$2\sqrt 2 $
Correct Answer: B
Explanation:
3x + 4y = 12$\sqrt 2 $
$ \Rightarrow $ y = $ - {{3x} \over 4} + 3\sqrt 2 $
is tangent to
$ \because $ equation of ellipse is ${{{x^2}} \over 4} + {{{y^2}} \over 8} = 1$
then it passes through $\left( {\sqrt 2 ,2} \right)$
2019
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2019 (Online) 12th April Morning Slot
If the normal to the ellipse 3x2
+ 4y2
= 12 at a point P on it is parallel to the line, 2x + y = 4 and the tangent
to the ellipse at P passes through Q(4,4) then PQ is equal to :
A.
${{\sqrt {61} } \over 2}$
B.
${{\sqrt {221} } \over 2}$
C.
${{\sqrt {157} } \over 2}$
D.
${{5\sqrt 5 } \over 2}$
Correct Answer: D
Explanation:
Equation of ellipse is ${{{x^2}} \over 4} + {{{y^2}} \over 3} = 1$
Normal at P(2 cos $\theta $, $\sqrt 3 \sin \theta $) is 2x sin$\theta $ - $\sqrt 3 y\,cos\theta $ = sin $\theta $ cos $\theta $ as the normal is parallel to 2x + y = 4
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2019 (Online) 10th April Evening Slot
The tangent and normal to the ellipse 3x2
+ 5y2
= 32 at the point P(2, 2) meet the x-axis at Q and R,
respectively. Then the area (in sq. units) of the triangle PQR is :
$ \therefore $ Area = ${1 \over 2}(QR) \times 2 = QR = {{68} \over {15}}$
2019
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2019 (Online) 10th April Morning Slot
If the line x – 2y = 12 is tangent to the ellipse ${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$ at the point $\left( {3, - {9 \over 2}} \right)$ , then the length of the latus
rectum of the ellipse is :
A.
5
B.
9
C.
$8\sqrt 3 $
D.
$12\sqrt 2 $
Correct Answer: B
Explanation:
Equation of tangent at $\left( {3, - {9 \over 2}} \right)$ to ${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$ is
${{3x} \over {{a^2}}} - {{{y^9}} \over {2{b^2}}} = 1$ which is equivalent to x – 2y = 12
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2019 (Online) 9th April Evening Slot
If the tangent to the parabola y2 = x at a point
($\alpha $, $\beta $), ($\beta $ > 0) is also a tangent to the ellipse,
x2 + 2y2 = 1, then $\alpha $ is equal to :
A.
$\sqrt 2 + 1$
B.
$\sqrt 2 - 1$
C.
$2\sqrt 2 + 1$
D.
$2\sqrt 2 - 1$
Correct Answer: A
Explanation:
Point P($\alpha $, $\beta $) is on the parabola y2 = x
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2019 (Online) 8th April Evening Slot
In an ellipse, with centre at the origin, if the
difference of the lengths of major axis and minor
axis is 10 and one of the foci is at (0,5$\sqrt 3$), then
the length of its latus rectum is :
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2019 (Online) 12th January Evening Slot
Let S and S' be the foci of an ellipse and B be any one of the extremities of its minor axis. If $\Delta $S'BS is a right angled triangle with right angle at B and area ($\Delta $S'BS) = 8 sq. units, then the length of a latus rectum of the ellipse is :
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2019 (Online) 11th January Evening Slot
Let the length of the latus rectum of an ellipse with its major axis along x-axis and centre at the origin, be 8. If the distance between the foci of this ellipse is equal to the length of its minor axis, then which one of the following points lies on it?
A.
$\left( {4\sqrt 2 ,2\sqrt 3 } \right)$
B.
$\left( {4\sqrt 3 ,2\sqrt 3 } \right)$
C.
$\left( {4\sqrt 3 ,2\sqrt 2 } \right)$
D.
$\left( {4\sqrt 2 ,2\sqrt 2 } \right)$
Correct Answer: C
Explanation:
${{2{b^2}} \over a} = 8$ and 2ae $=$ 2b
$ \Rightarrow $ ${b \over a}$ = e and 1 $-$ e2 = e2 $ \Rightarrow $ e $=$ ${1 \over {\sqrt 2 }}$
$ \Rightarrow $ b = 4$\sqrt 2 $ and a $=$ 8
So equation of ellipse is ${{{x^2}} \over {64}} + {{{y^2}} \over {32}} = 1$
2019
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2019 (Online) 11th January Morning Slot
If tangents are drawn to the ellipse x2 + 2y2 = 2 at all points on the ellipse other than its four vertices then the mid points of the tangents intercepted between the coordinate axes lie on the curve :
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2018 (Online) 16th April Morning Slot
If the length of the latus rectum of an ellipse is 4 units and the distance between a focus an its nearest vertex on the major axis is ${3 \over 2}$ units, then its eccentricity is :
A.
${1 \over 2}$
B.
${1 \over 3}$
C.
${2 \over 3}$
D.
${1 \over 9}$
Correct Answer: B
Explanation:
If the cordinate of focus and vertex are (ae, 0) and (a, 0) respectively,
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2017 (Online) 9th April Morning Slot
The eccentricity of an ellipse having centre at the origin, axes along the co-ordinate
axes and passing through the points (4, −1) and (−2, 2) is :
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2017 (Online) 8th April Morning Slot
Consider an ellipse, whose center is at the origin and its major axis is along the x-axis. If its eccentricity is ${3 \over 5}$ and the distance between its foci is 6, then the area (in sq. units) of the quadrilatateral inscribed in the ellipse, with the vertices as the vertices of the ellipse, is :
A.
8
B.
32
C.
80
D.
40
Correct Answer: D
Explanation:
e = 3/5 & 2ae = 6 $ \Rightarrow $ a = 5
$ \because $ b2 = a2 (1 $-$ e2)
$ \Rightarrow $ b2 = 25(1 $-$ 9/25)
$ \Rightarrow $ b = 4
$ \therefore $ Area of required quadrilateral
= 4(1/2 ab) = 2ab = 40
2017
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2017 (Offline)
The eccentricity of an ellipse whose centre is at the origin is ${1 \over 2}$. If one of its directrices is x = – 4, then the
equation of the normal to it at $\left( {1,{3 \over 2}} \right)$ is :
A.
2y – x = 2
B.
4x – 2y = 1
C.
4x + 2y = 7
D.
x + 2y = 4
Correct Answer: B
Explanation:
Given e = ${1 \over 2}$ and ${a \over e}$ = 4
$ \therefore $ $a$ = 2
We have b2 = $a$2 (1 – e2) = $4\left( {1 - {1 \over 4}} \right)$ = 3
$ \therefore $ Equation of ellipse is
${{{x^2}} \over 4} + {{{y^2}} \over 3} = 1$
Now, the equation of normal at $\left( {1,{3 \over 2}} \right)$ is
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2016 (Online) 9th April Morning Slot
If the tangent at a point on the ellipse ${{{x^2}} \over {27}} + {{{y^2}} \over 3} = 1$ meets the coordinate axes at A and B, and O is the origin, then the
minimum area (in sq. units) of the triangle OAB is :
$\Delta $ = will be minimum when sin 2$\theta $ = 1
$\Delta $min = 9
2015
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2015 (Offline)
The area (in sq. units) of the quadrilateral formed by the tangents at the end points of the latera recta to the ellipse ${{{x^2}} \over 9} + {{{y^2}} \over 5} = 1$, is :
A.
${{27 \over 2}}$
B.
$27$
C.
${{27 \over 4}}$
D.
$18$
Correct Answer: B
Explanation:
The end point of latus rectum of ellipse
${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$ in first quadrant is
$\left( {ae,{{{b^2}} \over a}} \right)$ and the tangent at this point
intersects $x$-axis at $\left( {{a \over e},0} \right)$ and
$y$-axis at $(0,a).$
The given ellipse is ${{x{}^2} \over 9} + {{{y^2}} \over 5} = 1$
An ellipse is drawn by taking a diameter of thec circle ${\left( {x - 1} \right)^2} + {y^2} = 1$ as its semi-minor axis and a diameter of the circle ${x^2} + {\left( {y - 2} \right)^2} = 4$ is semi-major axis. If the centre of the ellipse is at the origin and its axes are the coordinate axes, then the equation of the ellipse is :
A.
$4{x^2} + {y^2} = 4$
B.
${x^2} + 4{y^2} = 8$
C.
$4{x^2} + {y^2} = 8$
D.
${x^2} + 4{y^2} = 16$
Correct Answer: D
Explanation:
Equation of circle is ${\left( {x - 1} \right)^2} + {y^2} = 1$
$ \Rightarrow $ radius $=1$ and diameter $=2$
$\therefore$ Length of semi-minor axis is $2.$
Equation of circle is ${x^2} + {\left( {y - 2} \right)^2} = 4 = {\left( 2 \right)^2}$
The ellipse ${x^2} + 4{y^2} = 4$ is inscribed in a rectangle aligned with the coordinate axex, which in turn is inscribed in another ellipse that passes through the point $(4,0)$. Then the equation of the ellipse is :
A.
${x^2} + 12{y^2} = 16$
B.
$4{x^2} + 48{y^2} = 48$
C.
$4{x^2} + 64{y^2} = 48$
D.
${x^2} + 16{y^2} = 16$
Correct Answer: A
Explanation:
The given ellipse is ${{{x^2}} \over 4} + {{{y^2}} \over 1} = 1$
So $A=(2,0)$ and $B = \left( {0,1} \right)$
If $PQRS$ is the rectangular in which it is inscribed, then
$P = \left( {2,1} \right).$
Let ${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$
be the ellipse circumscribing the rectangular $PQRS$.
A focus of an ellipse is at the origin. The directrix is the line $x=4$ and the eccentricity is ${{1 \over 2}}$. Then the length of the semi-major axis is :
The eccentricity of an ellipse, with its centre at the origin, is ${1 \over 2}$. If one of the directrices is $x=4$, then the equation of the ellipse is :
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2026 (Online) 24th January Evening Shift
Let $(h, k)$ lie on the circle $\mathrm{C}: x^2+y^2=4$ and the point $(2 h+1,3 k+2)$ lie on an ellipse with eccentricity $e$. Then the value of $\frac{5}{e^2}$ is equal to $\_\_\_\_$ .
Correct Answer: 9
Explanation:
$(\mathrm{h}, \mathrm{k})$ lie on circle $x^2+y^2=1$.
so, $h^2+k^2=1--$ (1)
$(2 h+1), 3 k+2)$ lies on ellipse This means $x=2 h+1$ and $y=3 k+2$ lies on ellipse
Substitute $\mathrm{h}=\frac{x-1}{2}$ and $\mathrm{k}=\frac{y-2}{3}$ in equation (1)
This is shifted ellipse of the form $\frac{\left(x-x_1\right)^2}{-2}+\frac{\left(y-y_1\right)^2}{12}=1 \mathrm{a}=2, \mathrm{~b}=3$
eccentricity e $=\sqrt{1-\frac{a^2}{b^2}}=\sqrt{1-\frac{4}{9}}=\sqrt{\frac{5}{9}}$
$e^2=\frac{5}{9} \Longrightarrow \frac{5}{e^2}=9$ Ans
2025
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2025 (Online) 28th January Morning Shift
Let $\mathrm{E}_1: \frac{x^2}{9}+\frac{y^2}{4}=1$ be an ellipse. Ellipses $\mathrm{E}_{\mathrm{i}}$ 's are constructed such that their centres and eccentricities are same as that of $\mathrm{E}_1$, and the length of minor axis of $\mathrm{E}_{\mathrm{i}}$ is the length of major axis of $E_{i+1}(i \geq 1)$. If $A_i$ is the area of the ellipse $E_i$, then $\frac{5}{\pi}\left(\sum\limits_{i=1}^{\infty} A_i\right)$, is equal to _______.