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

NTA Abhyas 2020 · Mathematics · Parabola

Let lx - 2y = 1 intersects the parabola y ^ 2 = 4ax at points P and...

Let lx−2y=1lx - 2y = 1 intersects the parabola y2=4ax{y}^{2} = 4ax at points PP and Q.Q . If PSPS and QSQS meet the parabola again at RR and TT respectively (where SS is the focus of y2=4ax{y}^{2} = 4ax ), then the slope of RTRT is equal to

Correct answer

0.5

Step-by-step solution

Let, P=(t12,2t1)P = \left({t}_{1}^{2}, 2{t}_{1}\right) and Q = \left({t}_{2}^{2}, 2{t}_{2}\right)<br/>Equationof<br />Equation ofPSis(t1+t2)y=2x+2t1t2is \left({t}_{1} + {t}_{2}\right)y = 2x + 2{t}_{1}{t}_{2}
⇒⇒ lines −2x+(t1+t2)y=2t1t2- 2x + \left({t}_{1} + {t}_{2}\right)y = 2{t}_{1}{t}_{2} and lx−2y=1lx - 2y = 1 are same
⇒−2l=t1+t2−2=2t1t21⇒ \dfrac{- 2}{l} = \dfrac{{t}_{1} + {t}_{2}}{- 2} = \dfrac{2{t}_{1}{t}_{2}}{1}
⇒t1+t2=4l and t1t2=−1l⇒ {t}_{1} + {t}_{2} = \dfrac{4}{l} \text{ and } {t}_{1}{t}_{2} = \dfrac{- 1}{l}
Now, R=(1t12,−2t1)R = \left(\dfrac{1}{{t}_{1}^{2}}, \dfrac{- 2}{{t}_{1}}\right) and T=(1t22,−2t2)T = \left(\dfrac{1}{{t}_{2}^{2}}, \dfrac{- 2}{{t}_{2}}\right)
The slope of RTRT is −2t2+2t11t22−1t12=−2(t1−t2)t1t2(t1−t2)(t1+t2)t12t22\dfrac{\dfrac{- 2}{{t}_{2}} + \dfrac{2}{{t}_{1}}}{\dfrac{1}{{t}_{2}^{2}} - \dfrac{1}{{t}_{1}^{2}}} = \dfrac{- \dfrac{2\left({t}_{1} - {t}_{2}\right)}{{t}_{1}{t}_{2}}}{\dfrac{\left({t}_{1} - {t}_{2}\right)\left({t}_{1} + {t}_{2}\right)}{{t}_{1}^{2}{t}_{2}^{2}}}
=−2t1t2t1+t2=2l4l=12=0.5= \dfrac{- 2{t}_{1}{t}_{2}}{{t}_{1} + {t}_{2}} = \dfrac{\dfrac{2}{l}}{\dfrac{4}{l}} = \dfrac{1}{2} = 0 . 5
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