Question

Find the equation of the normal to the curve x2 = 4y which passes through the point(-1,4)
(CBSE 2019)​

Answer 1

Answer:

From the equation of the curve x

2

=4y,

we get

dy/dx=x/2. If m be the slope of the normal to x

2

=4y

then m=

(

dx

dy

)

−1

=−

x

2

∴x=

m

−2

and y=

4

x

2

=1/m

2

Thus normal is

y−

m

2

1

=m(x+

m

2

).

If it passes through the point (1,2),

then ∴2−

m

2

1

=m(1+

m

2

)=m+2∴m

3

=−1

orm=−1

∴Required normal is,

(y−2)=−1(x−1)

⇒x+y=3

Answer 2

Answer:

The equation of the normal to the given curve is $x+y=3$.

Step-by-step explanation:

It is asked to find the equation of the normal to the curve $x^{2} =4y$ which passes through the point (-1,4).

The line perpendicular to the tangent at the point of contact is called the normal to the parabola at that point.

The tangent is a straight line which just touches the curve at a given point.

$x^{2} =4y$

Differentiating both sides.

$2x\frac{dx}{dy}=4$$-\frac{2}{x} =\frac{ \frac{x^{2} }{4}-4}{-\frac{2}{x} -(-1)}$

Slope of the normal is $\frac{-1}{\frac{dy}{dx} } =-\frac{dx}{dy}$

Slope of normal is $-\frac{2}{x}$.

We know that point of contact is $(x,y)= (\frac{2a}{m} ,\frac{a}{m^{2} } )$

Here the value of a is 1.

So the point of contact will be $(x,y)= (\frac{2}{m} ,\frac{1}{m^{2} } )$

$x=\frac{2}{m}=-\frac{2}{x}$

$y=\frac{1}{m^{2} }=\frac{x^{2} }{4}$

Equation of normal is $y-\frac{1}{m^{2} } =m(x+\frac{2}{m} )$

Substituting (-1,4)

$4-\frac{1}{m^{2} } =m(-1+\frac{2}{m} )$

$4m^{2} -1=-m^{3} +2m^{2}$

$m^{3} +2m^{2}-1=0$

Solving this we get m=-1 as root.

Equation of normal is $y-4=-1(x+1)$

$y-4=-x-1$

$x+y=3$

To know more about normal go to the following link.

https://brainly.in/question/49716955

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https://brainly.in/question/10943031

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