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Parabola question

2010 · Shift 0 · Q41
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Parabola question

2010 · Shift 0 · Q41

JEE MainMathematicsParabolaMCQ+4 / −1
If two tangents drawn from a point PPP to the parabola y2=4x{y^2} = 4xy2=4x are at right angles, then the locus of PPP is
  1. A
    2x+1=02x+1=02x+1=0
  2. B
    x=−1x=-1x=−1
  3. C
    2x−1=02x-1=02x−1=0
  4. D
    x=1x=1x=1
View written solutionFree

Correct answer: B

  1. Equation of tangent to the parabola

For the parabola y2=4x,y^2=4x,y2=4x, its standard form is y2=4axy^2=4axy2=4ax with a=1a=1a=1.

A tangent with slope mmm to y2=4axy^2=4axy2=4ax is y=mx+am.y=mx+\frac{a}{m}.y=mx+ma​. Here a=1a=1a=1, so the tangent is y=mx+1m.y=mx+\frac{1}{m}.y=mx+m1​.

  1. Condition that the tangent passes through point P(h,k)P(h,k)P(h,k)

If this tangent passes through P(h,k)P(h,k)P(h,k), then k=mh+1m.k=mh+\frac{1}{m}.k=mh+m1​. Multiplying by mmm, hm2−km+1=0.hm^2-km+1=0.hm2−km+1=0.

Thus, the slopes of the two tangents from P(h,k)P(h,k)P(h,k) are the roots of hm2−km+1=0.hm^2-km+1=0.hm2−km+1=0. Let these slopes be m1m_1m1​ and m2m_2m2​.

  1. Condition for the tangents to be perpendicular

If the tangents are at right angles, then m1m2=−1.m_1m_2=-1.m1​m2​=−1.

From the quadratic equation, m1m2=1h.m_1m_2=\frac{1}{h}.m1​m2​=h1​. So, 1h=−1  ⟹  h=−1.\frac{1}{h}=-1 \implies h=-1.h1​=−1⟹h=−1.

Therefore, the locus of P(h,k)P(h,k)P(h,k) is x=−1.x=-1.x=−1.

  1. Check with options
  • A: 2x+1=0⇒x=−122x+1=0 \Rightarrow x=-\frac122x+1=0⇒x=−21​ ❌
  • B: x=−1x=-1x=−1 ✅
  • C: 2x−1=0⇒x=122x-1=0 \Rightarrow x=\frac122x−1=0⇒x=21​ ❌
  • D: x=1x=1x=1 ❌

Hence, the correct option is B.

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