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`        If y=m1x+c and y=m2x+c are two tangents to the parabola y^2+4a(x+a)=0, thena) m1+m2=0b) 1+ m1+m2=0c) m1m2-1=0d) 1+m1m2=0`
one year ago

```							lets consider a general line y=mx+c is a tangent to parabola y^2+4a(x+a)=0then (mx+c)^2+4a(x+a)=0 orm^2x^2+2x(2a+mc)+c^2+4a^2=0 has just one solution or discriminant is zero.(2a+mc)^2=m^2(c^2+4a^2)or 4a^2m^2 – 4acm – 4a^2=0or a^2m^2 – acm – a^2=0roots of the above eqn are m1 and m2. hence product of roots is-m1m2= – 4a^2/4a^2= – 1or m1m2+1=0
```
one year ago
```							Dear student y = m1( x+a) + a/ m1also y = m2 (x +a) + a/ m2 nowmake constant part equal a(1+ m1 ^2 )/m1 = a(1+ m2 ^2 )/m2 m1 + m1 * m2 ^2 = m2 + m2 * m1^2 m1- m2 = m1 m2 (m1 – m2) hence m1 m2 = 1
```
one year ago
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