Q. 11

# Find a point on t

Given:

The curve is y = 3x2 – 9x + 8

Differentiating the above w.r.t x

y = 3x2 – 9x + 8

= 23x2 – 1 – 9 + 0

= 6x – 9 ...(1)

Since, the tangent are equally inclined with axes

i.e, or

= The Slope of the tangent = tan

= tan() or tan()

= 1or – 1 ...(2)

tan() = 1

From (1) & (2),we get,

6x – 9 = 1 0r 6x – 9 = – 1

6x = 10 0r 6x = 8

x = or x =

x = or x =

Substituting x = or x = in y = 3x2 – 9x + 8,we get,

When x =

y = 3()2 – 9() + 8

y = 3() – () + 8

y = () – () + 8

taking LCM = 9

y = ()

y = ()

y = ()

y = ()

when x =

y = 3()2 – 9() + 8

y = 3() – () + 8

y = () – () + 8

taking LCM = 9

y = ()

y = ()

y = ()

y = ()

Thus, the required point is (,) & (,)

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