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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

2080 Points
3 years ago
lets consider a general line y=mx+c is a tangent to parabola y^2+4a(x+a)=0
then (mx+c)^2+4a(x+a)=0 or
m^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=0
or a^2m^2 – acm – a^2=0
roots of the above eqn are m1 and m2. hence product of roots is-
m1m2= – 4a^2/4a^2= – 1
or m1m2+1=0
Arun
25763 Points
3 years ago
Dear student

y = m1( x+a) + a/ m1
also

y = m2 (x +a) + a/ m2

now
make 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