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Question: Three normal to \[{{y}^{2}}=4x\] pass through the point (15, 12). If one of the normal is given by \...

Three normal to y2=4x{{y}^{2}}=4x pass through the point (15, 12). If one of the normal is given by y=x−3y=x-3 , find the equations of the others
A. y=−4x+72,y=3x−33y=-4x+72,y=3x-33
B. y=−3x+57,y=2x−18y=-3x+57,y=2x-18
C. y=−5x+97,y=−2x+42y=-5x+97,y=-2x+42
D. None of these

Explanation

Solution

In the given question we have been asked to find the equations of the normal to the given parabola y2=4x{{y}^{2}}=4x . As we know that the equation of the normal to the given parabola at any point‘t’ is y+tx=2t+t3y+tx=2t+{{t}^{3}} and it is given that the parabola passes through the point (15, 12), thus this point will satisfy the equation. Hence by putting the values we will simplify the equation and get different values of‘t’. By putting these value of‘t’ in the equation of the normal, we will get the remaining equations of the two normal.

Complete step by step solution:
We have given that,
y2=4x{{y}^{2}}=4x
As we know that,
Equation of the normal to the given parabola at any point ‘t’ is given by;
y+tx=2t+t3y+tx=2t+{{t}^{3}}
We have given that,
Three normal to y2=4x{{y}^{2}}=4x pass through the point (15, 12).
Thus, the point (x, y) = (15, 12) satisfies the equation of the normal i.e. y+tx=2t+t3y+tx=2t+{{t}^{3}} .
So,
Putting the value of (x, y) = (15, 12) in the equation, we get
⇒12+15t=2t+t3\Rightarrow 12+15t=2t+{{t}^{3}}
Subtracting 15t form both the sides of the equation, we get
⇒12=2t+t3−15t\Rightarrow 12=2t+{{t}^{3}}-15t
Combining the like terms, we get
⇒12=t3−13t\Rightarrow 12={{t}^{3}}-13t
Subtracting 12 from both the sides of the equation, we get
⇒t3−13t−12=0\Rightarrow {{t}^{3}}-13t-12=0
By long division
When we divide t3−13t−12{{t}^{3}}-13t-12 by t+1t+1 , we will get t2−t−12{{t}^{2}}-t-12
Thus,
⇒(t+1)(t2−t−12)=0\Rightarrow \left( t+1 \right)\left( {{t}^{2}}-t-12 \right)=0
Solving the quadratic equation by splitting the middle term, we get
⇒(t+1)(t2−4t+3t−12)=0\Rightarrow \left( t+1 \right)\left( {{t}^{2}}-4t+3t-12 \right)=0
Taking out the common factors, we will get
⇒(t+1)(t(t−4)+3(t−4))=0\Rightarrow \left( t+1 \right)\left( t\left( t-4 \right)+3\left( t-4 \right) \right)=0
⇒(t+1)(t−4)(t+3)=0\Rightarrow \left( t+1 \right)\left( t-4 \right)\left( t+3 \right)=0
Therefore,
⇒t=−1 or 4 or −3\Rightarrow t=-1\ or\ 4\ or\ -3
Now,
The three normal are;
Taking t = -1
⇒y+(−1)x=2(−1)+(−1)3⇒y−x=−2−1⇒y=−3+x\Rightarrow y+\left( -1 \right)x=2\left( -1 \right)+{{\left( -1 \right)}^{3}}\Rightarrow y-x=-2-1\Rightarrow y=-3+x
⇒y=x−3\Rightarrow y=x-3
Taking t = 4
⇒y+(4)x=2(4)+43⇒y+4x=8+64⇒y=72−4x\Rightarrow y+\left( 4 \right)x=2\left( 4 \right)+{{4}^{3}}\Rightarrow y+4x=8+64\Rightarrow y=72-4x
⇒y=−4x+72\Rightarrow y=-4x+72
Taking t = -3
⇒y+(−3)x=2(−3)+(−3)3⇒y−3x=−6−27⇒y=−33+3x\Rightarrow y+\left( -3 \right)x=2\left( -3 \right)+{{\left( -3 \right)}^{3}}\Rightarrow y-3x=-6-27\Rightarrow y=-33+3x
⇒y=3x−33\Rightarrow y=3x-33

So, the correct answer is Option A.

Note: While solving this question, students should need the concepts of parabola and the normal. The line that is perpendicular to the tangent of the given parabola at the point of their contact is known as the normal of that parabola. They should know that the Equation of the normal to the given parabola at any point ‘t’ is given by; y+tx=2t+t3y+tx=2t+{{t}^{3}} .