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Question

# The normal at P to the hyperbola x29−y21=1 meets the transverse axis AA′ at G and the conjugate axis BB′ at H. If CF is the perpendicular drawn from centre C of the hyperbola to the normal, then

A
PFPG=CB2
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B
PFPG=2CB2
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C
Locus of mid-point of GH is another hyperbola with eccentricity =10
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D
PFPH=CA2
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Solution

## The correct options are A PF⋅PG=CB2 C Locus of mid-point of GH is another hyperbola with eccentricity =√10 D PF⋅PH=CA2Given : x29−y21=1 C=(0,0), A=(3,0), B=(0,1) Let P=(3secθ,tanθ) Equation of normal in parametric form is axsecθ+bytanθ=a2+b2⇒3xcosθ+ycotθ=10 ∴G≡(10secθ3,0),H≡(0,10tanθ) PG=√sec2θ9+tan2θ⇒PG=√1+9sin2θ3|cosθ|PH=√9sec2θ+81tan2θ⇒PH=3√1+9sin2θ|cosθ| Now, the equation of CF is xcotθ−3ycosθ=0 Distance from P PF=∣∣∣3sinθ−3sinθ∣∣∣√cot2θ+9cos2θ⇒PF=3cos2θ√cos2θ+9sin2θcos2θ⇒PF=3|cosθ|√1+9sin2θPF⋅PG=1=CB2PF⋅PH=9=CA2 Let the midpoint of G and H be Q(h,k), then (h,k)=(5secθ3,5tanθ)sec2θ−tan2θ=1⇒(3h5)2−(k5)2=1 Therefore, the required locus is x2259−y225=1 This is a hyperbola, whose eccentricity =  ⎷1+25259=√10

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