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The Line 2x − Y + 4 = 0 Cuts the Parabola Y2 = 8x in P and Q. the Mid-point of Pq is - Mathematics

MCQ

The line 2x − y + 4 = 0 cuts the parabola y2 = 8x in P and Q. The mid-point of PQ is

Options

  •  (1, 2) 

  •  (1, −2) 

  •  (−1, 2) 

  •  (−1, −2) 

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Solution

(−1, 2) 

Let the coordinates of P and Q be \[\left( a {t_1}^2 , 2a t_1 \right)\] and \[\left( a {t_2}^2 , 2a t_2 \right)\]  respectively.
Slope of PQ = \[\frac{2a t_2 - 2a t_1}{a {t_2}^2 - a {t_1}^2}\]             ......(1) 

But, the slope of PQ is equal to the slope of 2x − y + 4 = 0.
∴ Slope of PQ = \[\frac{- 2}{- 1} = 2\]

From (1), \[\frac{2a t_2 - 2a t_1}{a {t_2}^2 - a {t_1}^2} = 2\]                        .....(2) 

Putting 4a = 8,
a = 2 

∴ Focus of the given parabola = (a, 0) = \[\left( 2, 0 \right)\] 

Using equation (2):

\[\frac{4\left( t_2 - t_1 \right)}{2\left( {t_2}^2 - {t_1}^2 \right)} = 2\] 

\[\frac{\left( t_2 - t_1 \right)}{\left( {t_2}^2 - {t_1}^2 \right)} = 1\] 

\[\frac{\left( t_2 - t_1 \right)}{\left( {t_2}^2 - {t_1}^2 \right)} = 1\] 

As, points P and Q lie on 2x-y+4=0 

\[\Rightarrow P(a {t_1}^2 , 2a t_1 ) or P(2 {t_1}^2 , 4 t_1 ) \text{  lie on line } 2x - y + 4 = 0\]
\[ \Rightarrow 2\left( 2 {t_1}^2 \right) - \left( 4 t_1 \right) + 4 = 0\]
\[ \Rightarrow {t_1}^2 - t_1 + 1 = 0 . . . (3)\]
\[\text{ Also }, Q(a {t_2}^2 , 2a t_2 ) or P(2 {t_2}^2 , 4 t_2 ) \text{ lie  on  line } 2x - y + 4 = 0\] 
\[ \Rightarrow 2\left( 2 {t_2}^2 \right) - \left( 4 t_2 \right) + 4 = 0\]
\[ \Rightarrow {t_2}^2 - t_2 + 1 = 0 . . . (4)\]
\[\text{ Adding } (3) \text{ and } (4), \text{ we get }, \]
\[ \Rightarrow {t_1}^2 - t_1 + 1 + {t_2}^2 - t_2 + 1 = 0\]
\[ \Rightarrow \left( {t_1}^2 + {t_2}^2 \right) - \left( t_1 + t_2 \right) + 2 = 0\]
\[ \Rightarrow \left( {t_1}^2 + {t_2}^2 \right) - 1 + 2 = 0 \left[ t_1 + t_2 = 1, \text{ proved above } \right]\]
\[ \Rightarrow \left( {t_1}^2 + {t_2}^2 \right) = - 1\]

Let \[\left( x_1 , y_1 \right)\]  be the mid-point of PQ.
Then, we have: \[y_1 = \frac{2a t_2 + 2a t_1}{2} = 2\left( t_1 + t_2 \right) = 2\]

\[x_1 = \frac{a {t_1}^2 + a {t_2}^2}{2} = {t_1}^2 + {t_2}^2 = - 1\] 

⇒ \[\left( x_1 , y_1 \right) = \left( - 1, 2 \right)\]

  Is there an error in this question or solution?
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APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 25 Parabola
Q 12 | Page 29
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