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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Mathematics
Chapter
Parabola

NTA Abhyas 2020 · Mathematics · Parabola

The parabolas C _ 1 : y ^ 2 = 4a(x - a) and C _ 2 : y ^ 2 = - 4a(x...

The parabolas C1:y2=4a(x−a){C}_{1}: {y}^{2} = 4a\left(x - a\right) and C2:y2=−4a(x−k){C}_{2}: {y}^{2} = - 4a\left(x - k\right) intersect at two distinct points AA and B.B . If the slope of the tangent at AA on C1{C}_{1} is same as the slope of the normal at BB on C2{C}_{2} , then the value of k is equal to

Options

  1. A3a3a
  2. B2a2a
  3. Caa
  4. D00

Correct answer

A. 3a3a

Step-by-step solution

Let, A=(p,q) and B=(p,−q)A = \left(p, q\right) \text{ and } B = \left(p, - q\right)
Now q2=4a(p−a) and q2=−4a(p−k){q}^{2} = 4a\left(p - a\right) \text{ and } {q}^{2} = - 4a\left(p - k\right) …(i)
⇒p−a=−p+k⇒p=a+k2⇒ p - a = - p + k ⇒ p = \dfrac{a + k}{2} …(ii)
Now for C1,dydx=2ay{C}_{1}, \dfrac{dy}{dx} = \dfrac{2a}{y}
{\left\dfrac{dy}{dx}\right]}_{\left(p, q\right)} = \dfrac{2a}{q} & {\left- \dfrac{dx}{dy}\right]}_{\left(p, - q\right)} = \dfrac{q}{2a}
⇒⇒ 2aq=q2a⇒q2=4a2\dfrac{2a}{q} = \dfrac{q}{2a} ⇒ {q}^{2} = 4{a}^{2} … (iii)
Putting values from (ii) & (iii) in equation (i), we get,
4a2=4a(a+k2−a)4{a}^{2} = 4a\left(\dfrac{a + k}{2} - a\right)⇒ a = \dfrac{k - a}{2} ⇒ k = 3a
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