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Miscellaneous Exercise 7(II) · Q128

Q.The tangent at point P on the parabola y2=4axy^2 = 4ax meets the y-axis in Q. If S is the focus, show that SP subtends a right angle at Q.

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Let P=(at2,2at)P=(at^2,2at). The tangent at PP is yt=x+at2yt=x+at^2. At x=0x=0: yt=at2⇒y=atyt=at^2 \Rightarrow y=at (for t≠0t\ne0), so Q=(0,at)Q=(0,at).

The focus is S=(a,0)S=(a,0).

QS⃗=S−Q=(a−0, 0−at)=(a,−at),QP⃗=P−Q=(at2−0, 2at−at)=(at2,at).\vec{QS}=S-Q=(a-0,\,0-at)=(a,-at),\qquad \vec{QP}=P-Q=(at^2-0,\,2at-at)=(at^2,at).

QS⃗⋅QP⃗=a(at2)+(−at)(at)=a2t2−a2t2=0.\vec{QS}\cdot\vec{QP}=a(at^2)+(-at)(at)=a^2t^2-a^2t^2=0. …

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