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1

Numerical

Suppose that the foci of the ellipse $${{{x^2}} \over 9} + {{{y^2}} \over 5} = 1$$ are $$\left( {{f_1},0} \right)$$ and $$\left( {{f_2},0} \right)$$ where $${{f_1} > 0}$$ and $${{f_2} < 0}$$. Let $${P_1}$$ and $${P_2}$$ 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 $${T_1}$$ be a tangent to $${P_1}$$ which passes through $$\left( 2{{f_2},0} \right)$$ and $${T_2}$$ be a tangent to $${P_2}$$ which passes through $$\left( {{f_1},0} \right)$$. If $${m_1}$$ is the slope of $${T_1}$$ and $${m_2}$$ is the slope of $${T_2}$$, then the value of $$\left( {{1 \over {m_1^2}} + m_2^2} \right)$$ is

Your Input ________

Correct Answer is **4**

$${e^2} = 1 - {{{b^2}} \over {{a^2}}} = 1 - {5 \over 9} = {4 \over 9}$$

The foci are ($$\pm$$ ae, 0) i.e. (2, 0) and ($$-$$2, 0).

The parabola P_{1} is $${y^2} = 8x$$ and P_{2} is $${y^2} = - 16x$$

As tangent with slope m_{1} to P_{1} passes through ($$-$$4, 0), we have

$$y = {m_1}x + {2 \over {{m_1}}}$$ giving $$0 = - 4{m_1} + {2 \over {{m_1}}}$$

i.e. $$4m_1^2 = 2 \Rightarrow m_1^2 = {1 \over 2}$$

Again for tangent with slope m_{2} to P_{2} passing through (2, 0), we have

$$y = {m_2}x - {4 \over {{m_2}}} \Rightarrow 0 = 2{m_2} - {4 \over {{m_2}}}$$

$$ \Rightarrow 2m_2^2 = 4$$ $$\therefore$$ $$m_2^2 = 2$$

Thus, $${1 \over {m_1^2}} + m_2^2 = 2 + 2 = 4$$

2

Numerical

Let the curve $$C$$ be the mirror image of the parabola $${y^2} = 4x$$ with respect to the line $$x+y+4=0$$. If $$A$$ and $$B$$ are the points of intersection of $$C$$ with the line $$y=-5$$, then the distance between $$A$$ and $$B$$ is

Your Input ________

Correct Answer is **4**

Let, P(t^{2}, 2t) be any point on the parabola y^{2} = 4x. C be the mirror image of the parabola y^{2} = 4x with respect to the line UV : x + y + 4 = 0.

The curve C cuts the line KL : y = $$-$$5 at A and B.

Let, B($$\alpha$$, $$\beta$$) be the image of the point P(t^{2}, 2t).

Clearly, PB $$\bot$$ UV and PQ = QB.

$$\therefore$$ $${{\alpha - {t^2}} \over {\beta - 2t}} \times ( - 1) = - 1$$

or, $$\alpha - {t^2} = \beta - 2t$$ ...... (1)

The point of intersection of the lines UV and KL is R.

Let us join P and R.

From $$\Delta$$PQR and $$\Delta$$BQR,

(i) BQ = PQ [$$\because$$ B is the image of P]

(ii) $$\angle$$PQR = $$\angle$$RQB = 90$$^\circ$$ [$$\because$$ PB $$\bot$$ UV]

(iii) QR common

$$\therefore$$ $$\Delta$$PQR $$ \cong $$ $$\Delta$$BQR [by SAS congruence criterion]

$$\therefore$$ $$\angle$$QRP = $$\angle$$BRQ [CPCT]

$$\because$$ slope of x + y + 4 = 0 is $$-$$1,

$$\therefore$$ $$\angle$$UTO = 135$$^\circ$$

$$\therefore$$ $$\angle$$OTR = 45$$^\circ$$

Again, X'X || KL and UV transversal.

$$\therefore$$ $$\angle$$OTR = $$\angle$$TRB = 45$$^\circ$$ $$\therefore$$ $$\angle$$BRQ = $$\angle$$QRP = 45$$^\circ$$

$$\therefore$$ $$\angle$$PRB = 90$$^\circ$$ $$\therefore$$ PR $$\bot$$ KL

$$\therefore$$ coordinates of R are (t^{2}, $$\beta$$).

$$\because$$ the point R lies on KL,

$$\therefore$$ $$\beta$$ = $$-$$5

Again, the point R lies on the straight line x + y + 4 = 0.

$$\therefore$$ t^{2} + $$\beta$$ + 4 = 0

or, t^{2} $$-$$ 5 + 4 = 0 [$$\because$$ $$\beta$$ = $$-$$5]

or, t^{2} = 1 or, t = $$\pm$$ 1

when t = 1, $$\beta$$ = $$-$$5, then (1) $$\Rightarrow$$ $$\alpha$$ $$-$$ 1 = $$-$$ 5 $$-$$ 2 or $$\alpha$$ = $$-$$6

when, t = $$-$$1, $$\beta$$ = $$-$$5, then (1) $$\Rightarrow$$ $$\alpha$$ $$-$$ 1 = $$-$$ 5 + 2 or, $$\alpha$$ = $$-$$2

So, the coordinates of A and B are ($$-$$6, $$-$$5) and ($$-$$2, $$-$$5) respectively.

$$\therefore$$ AB = 4 units

So, the distance between A and B is 4 units.

3

Numerical

If the normals of the parabola $${y^2} = 4x$$ drawn at the end points of its latus rectum are tangents to the circle $${\left( {x - 3} \right)^2} + {\left( {y + 2} \right)^2} = {r^2}$$, then the value of $${r^2}$$ is

Your Input ________

Correct Answer is **2**

4

Numerical

A vertical line passing through the point $$(h,0)$$ intersects the ellipse $${{{x^2}} \over 4} + {{{y^2}} \over 3} = 1$$ at the points $$P$$ and $$Q$$. Let the tangents to the ellipse at $$P$$ and $$Q$$ meet at the point $$R$$. If $$\Delta \left( h \right)$$$$=$$ area of the triangle $$PQR$$, $${{\Delta _1}}$$ $$ = \mathop {\max }\limits_{1/2 \le h \le 1} \Delta \left( h \right)$$ and $${{\Delta _2}}$$ $$ = \mathop {\min }\limits_{1/2 \le h \le 1} \Delta \left( h \right)$$, then $${8 \over {\sqrt 5 }}{\Delta _1} - 8{\Delta _2} = $$

Your Input ________

Correct Answer is **9**

On those following papers in Numerical

Number in Brackets after Paper Indicates No. of Questions

JEE Advanced 2021 Paper 2 Online (1)

JEE Advanced 2017 Paper 1 Offline (1)

JEE Advanced 2015 Paper 2 Offline (1)

JEE Advanced 2015 Paper 1 Offline (2)

JEE Advanced 2013 Paper 1 Offline (1)

IIT-JEE 2012 Paper 1 Offline (1)

IIT-JEE 2011 Paper 1 Offline (1)

IIT-JEE 2010 Paper 1 Offline (1)

Complex Numbers

Quadratic Equation and Inequalities

Permutations and Combinations

Mathematical Induction and Binomial Theorem

Sequences and Series

Matrices and Determinants

Vector Algebra and 3D Geometry

Probability

Trigonometric Functions & Equations

Properties of Triangle

Inverse Trigonometric Functions

Straight Lines and Pair of Straight Lines

Circle

Conic Sections

Functions

Limits, Continuity and Differentiability

Differentiation

Application of Derivatives

Indefinite Integrals

Definite Integrals and Applications of Integrals

Differential Equations