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`        If the normal subtends a right angle at the focus of the parabola y2=8x then its length is...? a)2√5  b)10√5  c)5√5  d)4√5`
3 years ago

## Answers : (1)

J Sampat Srivatsav
28 Points
```							let S=(2,0) be focus of the given parabola Let P(at12,2at1) and Q(at22,2at2) be the points at which the normal intersects the parabolaslope of SP=4t1/2(t12-1)=2t1/t12-1similarly,slope of SQ=2t2/t22-1product of slopes=-1(2t1)(2t2)/(t12-1)(t22-1)=-1t12t22+t12+t22+1+4t1t2=0t12t22+2t1t2+1=t12+t22-2t1t2(t1t2+1)=t1-t2t1t2+r2=t1-1t2(t1+1)=t1-1t2=t1-1/t1+1-t1-2/t1=t1-1/t1+1on solving t1=-2t2=-t1-2/t1t2=3P=(8,-8) ; Q=(18,12)PQ2=(18-8)2+(12-(-8))2=100+400=500PQ=10(5)1/2
```
2 years ago
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• 731 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions