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Ellipse question

2010 · Shift 2 · Q27
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  5. /2010 · Shift 2 · Q27

Ellipse question

2010 · Shift 2 · Q27

JEE AdvancedMathematicsEllipseMCQ+4 / −1
Tangents are drawn from the point P(3,4)P(3, 4)P(3,4) to the ellipse x29+y24=1{{{x^2}} \over 9} + {{{y^2}} \over 4} = 19x2​+4y2​=1 touching the ellipse at points AAA and BBB. The coordinates of AAA and BBB are
  1. A
    (3,0)(3,0)(3,0) and (0,2)(0,2)(0,2)
  2. B
    (−85,216115)\left( { - {8 \over 5},{{2\sqrt {161} } \over {15}}} \right)(−58​,152161​​) and (−95,85)\left( { - {9 \over 5},{8 \over 5}} \right)(−59​,58​)
  3. C
    (−85,216115)\left( { - {8 \over 5},{{2\sqrt {161} } \over {15}}} \right)(−58​,152161​​) and (0,2)(0,2)(0,2)
  4. D
    (3,0)(3,0)(3,0) and (−95,85)\left( { - {9 \over 5},{8 \over 5}} \right)(−59​,58​)
View written solutionFree

Correct answer: D

  1. Given ellipse and external point

The ellipse is

x29+y24=1\frac{x^2}{9}+\frac{y^2}{4}=19x2​+4y2​=1

with a2=9, b2=4a^2=9,\ b^2=4a2=9, b2=4 and the external point is

P(3,4).P(3,4).P(3,4).

We need the points of contact AAA and BBB of tangents drawn from PPP to the ellipse.

  1. Equation of tangent at a point on the ellipse

If (x1,y1)(x_1,y_1)(x1​,y1​) is a point on the ellipse, then the tangent at that point is

xx19+yy14=1.\frac{xx_1}{9}+\frac{yy_1}{4}=1.9xx1​​+4yy1​​=1.

Since this tangent passes through P(3,4)P(3,4)P(3,4), substituting (x,y)=(3,4)(x,y)=(3,4)(x,y)=(3,4) gives

3x19+4y14=1.\frac{3x_1}{9}+\frac{4y_1}{4}=1.93x1​​+44y1​​=1.

So,

x13+y1=1.\frac{x_1}{3}+y_1=1.3x1​​+y1​=1.

Hence the point of contact (x1,y1)(x_1,y_1)(x1​,y1​) must satisfy

y1=1−x13.y_1=1-\frac{x_1}{3}.y1​=1−3x1​​.

Also, since (x1,y1)(x_1,y_1)(x1​,y1​) lies on the ellipse,

x129+y124=1.\frac{x_1^2}{9}+\frac{y_1^2}{4}=1.9x12​​+4y12​​=1.
  1. Substitute into ellipse equation

Substitute y1=1−x13y_1=1-\frac{x_1}{3}y1​=1−3x1​​ into the ellipse:

x129+14(1−x13)2=1.\frac{x_1^2}{9}+\frac{1}{4}\left(1-\frac{x_1}{3}\right)^2=1.9x12​​+41​(1−3x1​​)2=1.

Expand:

x129+14(1−2x13+x129)=1.\frac{x_1^2}{9}+\frac{1}{4}\left(1-\frac{2x_1}{3}+\frac{x_1^2}{9}\right)=1.9x12​​+41​(1−32x1​​+9x12​​)=1.

Multiply through by 363636:

4x12+9(1−2x13+x129)=36.4x_1^2+9\left(1-\frac{2x_1}{3}+\frac{x_1^2}{9}\right)=36.4x12​+9(1−32x1​​+9x12​​)=36. 4x12+9−6x1+x12=36.4x_1^2+9-6x_1+x_1^2=36.4x12​+9−6x1​+x12​=36. 5x12−6x1−27=0.5x_1^2-6x_1-27=0.5x12​−6x1​−27=0.

Solve:

x1=6±36+54010=6±57610=6±2410.x_1=\frac{6\pm\sqrt{36+540}}{10}=\frac{6\pm\sqrt{576}}{10}=\frac{6\pm24}{10}.x1​=106±36+540​​=106±576​​=106±24​.

Thus,

x1=3orx1=−95.x_1=3 \quad \text{or} \quad x_1=-\frac{9}{5}.x1​=3orx1​=−59​.

Now find corresponding y1y_1y1​ values:

  • If x1=3x_1=3x1​=3,

    y1=1−33=0.y_1=1-\frac{3}{3}=0.y1​=1−33​=0.

    So one point is

    (3,0).(3,0).(3,0).

  • If x1=−95x_1=-\frac{9}{5}x1​=−59​,

    y1=1−−9/53=1+35=85.y_1=1-\frac{-9/5}{3}=1+\frac{3}{5}=\frac{8}{5}.y1​=1−3−9/5​=1+53​=58​.

    So the other point is

    (−95,85).\left(-\frac{9}{5},\frac{8}{5}\right).(−59​,58​).
  1. Points of contact

Therefore,

A,B=(3,0) and (−95,85).A,B=(3,0) \text{ and } \left(-\frac{9}{5},\frac{8}{5}\right).A,B=(3,0) and (−59​,58​).

This matches Option D.

  1. Comparison with stored correct answer

Stored correct answer: D

Our derived answer: D

So the stored answer is correct.

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