PYQ Vault
EbooksTutor

Common tangent to two conics using the condition of tangency

This type of question, asked 4 times: 2021 Paper 8, 2021 Paper 13, 27 Jul 2022, 28 Jul 2022. All 4 questions below, oldest first.

Try each one first, then open its answer.

  1. Question 1

    2021 Paper 8

    76
    Let LL be a tangent line to the parabola y2=4x−20y^{2}= 4x - 20 at (6,2)(6,2). If LL is also a tangent to the ellipse x22+y2b=1\frac{x^{2}}{2}+\frac{y^{2}}{b}= 1, then the value of bb is equal to:

    Tap an option to check your answer.

  2. Question 2

    2021 Paper 13

    54
    If P:y2=αx;L:2x+y=k LP:y^{2}=\alpha x;L:2x+y=k\ L is a tangent to x2−y2=3x^{2}-y^{2}= 3 and PP. Find ' α\alpha '

    Tap an option to check your answer.

  3. Question 3

    27 Jul 2022

    179
    A common tangent TT to the curves C1:x24+y29=1C_{1}:\frac{x^{2}}{4}+\frac{y^{2}}{9}= 1 and C2:x242−y2143=1C_{2}:\frac{x^{2}}{42}-\frac{y^{2}}{143}= 1 does not pass through the fourth quadrant. If TT touches C1C_{1} at (x1,y1)\left( x_{1},y_{1} \right) and C2C_{2} at (x2,y2)\left( x_{2},y_{2} \right), then ∣2x1+x2∣\left| 2x_{1}+x_{2} \right| is equal to

  4. Question 4

    28 Jul 2022

    87
    For the hyperbola H:x2−y2=1H:x^{2}-y^{2}= 1 and the ellipse E:x2a2+y2b2=1,a>b>0E:\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}= 1,a > b > 0, let the (1) eccentricity of EE be reciprocal of the eccentricity of HH, and (2) the line y=52x+Ky =\sqrt{\frac{5}{2}}x + K be a common tangent of EE and HH. Then 4(a2+b2)4\left( a^{2}+b^{2} \right) is equal to