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Byju's Answer
Standard XII
Mathematics
Equation of Normal at a Point (x,y) in Terms of f'(x)
The equation ...
Question
The equation of the normal to the curve 3x
2
− y
2
= 8 which is parallel to x + 3y = 8 is
(a) x + 3y = 8
(b) x + 3y + 8 = 0
(c) x + 3y ± 8 = 0
(d) x + 3y = 0
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Solution
(c) x + 3y ± 8 = 0
The slope of line x + 3y = 8 is
-
1
3
.
Since the normal is parallel to the given line, the equation of normal will be of the given form.
x
+
3
y
=
k
3
x
2
-
y
2
=
8
Let
x
1
,
y
1
be
the
point
of
intersection
of
the
two
curves
.
Then
,
x
1
+
3
y
1
=
k
.
.
.
1
3
x
1
2
-
y
1
2
=
8
.
.
.
2
Now
,
3
x
2
-
y
2
=
8
On
d
ifferentiating both sides w.r.t.
x
,
we
get
6
x
-
2
y
d
y
d
x
=
0
⇒
d
y
d
x
=
6
x
2
y
=
3
x
y
Slope of the tangent =
d
y
d
x
x
1
,
y
1
=
3
x
1
y
1
Slope of the normal
,
m
=
-
1
3
x
y
=
-
y
1
3
x
1
Given:
Slope of the normal = Slope of the given line
⇒
-
y
1
3
x
1
=
-
1
3
⇒
y
1
=
x
1
.
.
.
3
From (2), we get
3
x
1
2
-
x
1
2
=
8
⇒
2
x
1
2
=
8
⇒
x
1
2
=
4
⇒
x
1
=
±
2
Case 1:
When
x
1
=2
From (3), we get
y
1
=
x
1
=
2
∴
x
1
,
y
=
2
,
2
From (1), we get
2
+
3
2
=
k
⇒
2
+
6
=
k
⇒
k
=
8
∴
Equation of the normal from (1)
⇒
x
+
3
y
=
8
⇒
x
+
3
y
-
8
=
0
Case 2:
When
x
1
=-2
From (3), we get
y
1
=
x
1
=
-
2
,
∴
x
1
,
y
=
-
2
,
-
2
From (1), we get
-
2
+
3
-
2
=
k
⇒
-
2
-
6
=
k
⇒
k
=
-
8
∴
Equation of the normal from (1)
⇒
x
+
3
y
=
-
8
⇒
x
+
3
y
+
8
=
0
From both the cases, we get the equation of the normal as:
x
+
3
y
±
8
=
0
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0
Similar questions
Q.
Find the equation of the normal lines to the curve
3
x
2
−
y
2
=
8
which are parallel to the line
x
+
3
y
=
4.
Q.
The equation(s) of normal(s) to the curve
3
x
2
−
y
2
=
8
which is (are) parallel to the line x + 3y = 4 is (are)
