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Hint: To find the locus of point of intersection of tangents to the two parabolas

which are perpendicular to each other, write the equation of tangents in slope form and

solve them to find the locus of their point of intersection. Also, the product of slope of two

perpendicular lines is -1.

We have two parabolas \[{{y}^{2}}=4\left( x+1 \right)\]and\[{{y}^{2}}=8\left( x+2 \right)\].

To find the locus of points of intersection of perpendicular tangents to the two parabolas, we

will first write the equation of tangents to the parabolas in slope form.

We know that equation of tangent of a parabola of the form \[{{y}^{2}}=4a\left( x-b

\right)\]with slope\[m\] is \[y=m\left( x-b \right)+\dfrac{a}{m}\].

Let’s assume that the slope of tangents to one of the parabolas is \[m\].

We know that the product of slopes of two perpendicular lines is -1.

So, the slope of the other tangent is \[\dfrac{-1}{m}\].

For the parabola \[{{y}^{2}}=4\left( x+1 \right)\], we have \[a=1,b=-1\].

Substituting the above values in equation of tangent, we get \[y=m\left( x+1

\right)+\dfrac{1}{m}=mx+\dfrac{1}{m}+m\]. \[\left( 1 \right)\]

For the parabola \[{{y}^{2}}=8\left( x+2 \right)\], we have \[a=2,b=-2\].

Substituting the above values in equation of tangent, we get \[y=\dfrac{-1}{m}\left( x+2

\right)+\dfrac{2}{\dfrac{-1}{m}}=\dfrac{-x}{m}-\dfrac{2}{m}-2m\]. \[\left( 2

\right)\]

Subtracting equation \[\left( 1 \right)\] from equation \[\left( 2 \right)\], we get

\[mx+\dfrac{x}{m}+3m+\dfrac{3}{m}=0\].

\[\Rightarrow x\left( m+\dfrac{1}{m} \right)+3\left( m+\dfrac{1}{m} \right)=0\]

\[\Rightarrow x+3=0\]

We observe that the locus of points of intersection of perpendicular tangents is a straight

line. We can also find the point at which the tangent intersects the parabola. We just have to

substitute the equation of parabola in the equation of tangent and solve the equation to get

their point of intersection.

Hence, the correct answer is \[x+3=0\] which is option (c).

Note: We can also write the equation of tangents in the point form taking the locus as point

of intersection of two tangents and then substituting the product of slopes of two tangents

as -1.

which are perpendicular to each other, write the equation of tangents in slope form and

solve them to find the locus of their point of intersection. Also, the product of slope of two

perpendicular lines is -1.

We have two parabolas \[{{y}^{2}}=4\left( x+1 \right)\]and\[{{y}^{2}}=8\left( x+2 \right)\].

To find the locus of points of intersection of perpendicular tangents to the two parabolas, we

will first write the equation of tangents to the parabolas in slope form.

_{}We know that equation of tangent of a parabola of the form \[{{y}^{2}}=4a\left( x-b

\right)\]with slope\[m\] is \[y=m\left( x-b \right)+\dfrac{a}{m}\].

Let’s assume that the slope of tangents to one of the parabolas is \[m\].

We know that the product of slopes of two perpendicular lines is -1.

So, the slope of the other tangent is \[\dfrac{-1}{m}\].

For the parabola \[{{y}^{2}}=4\left( x+1 \right)\], we have \[a=1,b=-1\].

Substituting the above values in equation of tangent, we get \[y=m\left( x+1

\right)+\dfrac{1}{m}=mx+\dfrac{1}{m}+m\]. \[\left( 1 \right)\]

For the parabola \[{{y}^{2}}=8\left( x+2 \right)\], we have \[a=2,b=-2\].

Substituting the above values in equation of tangent, we get \[y=\dfrac{-1}{m}\left( x+2

\right)+\dfrac{2}{\dfrac{-1}{m}}=\dfrac{-x}{m}-\dfrac{2}{m}-2m\]. \[\left( 2

\right)\]

Subtracting equation \[\left( 1 \right)\] from equation \[\left( 2 \right)\], we get

\[mx+\dfrac{x}{m}+3m+\dfrac{3}{m}=0\].

\[\Rightarrow x\left( m+\dfrac{1}{m} \right)+3\left( m+\dfrac{1}{m} \right)=0\]

\[\Rightarrow x+3=0\]

We observe that the locus of points of intersection of perpendicular tangents is a straight

line. We can also find the point at which the tangent intersects the parabola. We just have to

substitute the equation of parabola in the equation of tangent and solve the equation to get

their point of intersection.

Hence, the correct answer is \[x+3=0\] which is option (c).

Note: We can also write the equation of tangents in the point form taking the locus as point

of intersection of two tangents and then substituting the product of slopes of two tangents

as -1.

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