Tangents are drawn from points on line \[x - y + 3 = 0\] to a parabola \[{y^2} = 8x\]. Then the variable chord of contact passes through a fixed point whose coordinates will be?
VerifiedHint: We use the method of substitution to find a point on the given line. Use the equation of chord of contact with respect to a given point of a parabola and substitute the values obtained from the point. Solve to get the value of coordinates.
Chord of contact with respect to point \[({x_1},{y_1})\] of a parabola \[{y^2} = 4ax\] is given
by \[y{y_1} = 2a(x + {x_1})\].
Complete step-by-step answer:
We are given the equation of line as \[x - y + 3 = 0\]
The point through which tangents are drawn to the parabola lies on the line. We use equation of line
to find the point.
Let us assume the value of x-coordinate as a variable, say \[x = k\].
Then substitute the value of x in the equation of line to find the y-coordinate.
\[ \Rightarrow k - y + 3 = 0\]
Shift y to RHS of the equation
\[ \Rightarrow k + 3 = y\]
So, the point on the line \[x - y + 3 = 0\] is \[(k,k + 3)\]. …………..… (1)
Now we are given an equation of parabola as \[{y^2} = 8x\].
Compare the equation of parabola with general equation of parabola i.e. \[{y^2} = 4ax\]
\[ \Rightarrow 4a = 8\]
\[ \Rightarrow a = 2\] ………..… (2)
Since we know that the equation of chord of contact is given by\[y{y_1} = 2a(x + {x_1})\].
Substitute the value of point from equation (1) and value of a from equation (2)
Put \[{x_1} = k,{y_1} = k + 3,a = 2\]
\[ \Rightarrow y(k + 3) = 2 \times 2(x + k)\]
\[ \Rightarrow yk + 3y = 4x + 4k\]
Bring all the terms to LHS of the equation
\[ \Rightarrow yk + 3y - 4x - 4k = 0\]
Collect the terms having value ‘k’ common
\[ \Rightarrow 3y - 4x + (yk - 4k) = 0\]
Take ‘k’ common from second bracket
\[ \Rightarrow (3y - 4x) + k(y - 4) = 0\]
Since ‘k’ is a constant value, therefore if we can take \[(3y - 4x) = {P_1};(y - 4) = {P_2}\]as two straight lines
\[ \Rightarrow {P_1} + k{P_2} = 0\]
Now the point will pass through the intersection of straight lines \[{P_1},{P_2}\]
Put \[{P_2} = 0\]
\[ \Rightarrow y - 4 = 0\]
Shifting the value of constant term to RHS
\[ \Rightarrow y = 4\]
Put \[{P_1} = 0\]
\[ \Rightarrow 3y - 4x = 0\]
Substitute the value of ‘y’ as 4
\[ \Rightarrow 3 \times 4 - 4x = 0\]
\[ \Rightarrow 12 - 4x = 0\]
Shifting the value of constant term to RHS
\[ \Rightarrow - 4x = - 12\]
Divide both sides by -4
\[ \Rightarrow \dfrac{{ - 4x}}{{ - 4}} = \dfrac{{ - 12}}{{ - 4}}\]
Cancel same terms from numerator and denominator on both sides of the equation
\[ \Rightarrow x = 3\]
\[\therefore \]The point becomes \[(3,4)\]
Note: Chord joining the two points of contact from the tangents to the parabola from same external points is called the chord of contact. It is of the form\[T = 0\]. PQ is the chord of contact.
Students might make mistake in the solution as they take the value of ‘a’ as 8 when they see
the equation of parabola, which is wrong. Keep in mind we compare the equation with the general equation and then find the value of ‘a’.
