The shortest distance between the line y = x and the curve y² = x - 2 is ?
Answers
Answer:
do same as this:-
Question:-
ΔABC is an isosceles triangle in which ∠C= 90° . If AC = 6 cm, then AB=
A. 6√2 cm
B. 6 cm
C. 2√6 cm
D. 4√2cm
Answer:-
In triangle ABC
Let tge altitude be AD on BC
AC=25cm
AB=25cm
BC=14cm
Therefore DC= 1/2*BC
=1/2*14
=7
By Pythagoras theorem,
AD^2+CD^2=AC^2
AD^2+7^2=25^2
AD^2+49=625
AD^2=625-49
AD^2=576
AD=24
Therefore altitude from A on BC is 24 cm
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Since solution to y = x and y^2 = x - 2 doesn't exist, the shortest distance between them must be perpendicular to the curve and line. It means, both would have same slope at such condition.
For slope of y^2 = x - 2, diff. wrt. x :
=> 2yy' = 1 => y' = 1/(2y)
For slope of line y = x, use y = mx + c
=> m = 1
As both should have same slope, m = y'
=> 1 = 1/2y => y = 1/2
Therefore, when y = 1/2 (on the curve)
(1/2)^2 = x - 2 => 9/4 = x
Hence the point where perpendicular meets on curve is (9/4, 1/2).
Now we need to find the point on line that is perpendicular to curve(or line with (9/4, 1/2) and slope = - 1/1).
Eqⁿ of perpendicular is
=> y - 1/2 = -1 (x - 9/4)
=> 4y + 4x - 11 = 0
Now, the lines y = x and 4y + 4x - 11 = 0 meet at :
(x, y) = (11/8, 11/8).
We have a point on line(11/8, 11/8) and perpendicularly a point(9/4, 1/2) on the curve. Using distance formula:
=> √(11/8 - 9/4)² + (11/8 - 1/2)²
=> 7√2/8
