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`        The normal to the curve 5x^5-10x^3+X+2y+6=0  at the point p(0,-3) is tangent to the curve at some other points.find those points??`
one year ago

```							Differentiating curve equation25x4 – 30x2 + 1 + 2m =02m= 30x2 - 25x4 – 1m =15x2 – (25/2)x4 – 1/2m(0,-3) = -1/2So slope of normal at p(0, -3) = -1/m(0,-3) =2It is also tangent to curve at some points (x,y) thenm(x,y) = 2 =15x2 – (25/2)x4 – ½So 3 = 30x2 – 25x425x4 - 30x2 +3 = 0Solve this equation for value of x and then put value of x in equation of curve to find value of y.
```
one year ago
```							Dear Aayushi Let us assume the question as: The normal to the curve 5x5 - 10x3 + x + 2y + 6 = 0 at point (0, -3) is tangent to the curve at some other point. Find the point ?To solve this, we first have to find the equation of normal to the given the given curve at the point (0, -3). For that, we need to differentiate the equation of the curve and find the gradient. The gradient of the given curve will be:   25 x⁴ - 30 x² +1 + 2dy/dx = 0 Now value of gradient at (0 , -3)dy/dx = -1/2 Let as assume the value as m1 = -1/2We also know that slope of the normal will be: m2 = (-1 / m1). Hence, m2 = 2.So, equation of the normal can be found by using point slope form:(y - y1) = m2 (x - x1)(y - (-3)) = 2 (x - 0)y + 3 = 2xy = 2x - 3 ---------- (i)According to the question, equation (i) is also tangent to the given curve. We can plug in the value of y from equation (i) into the equation of the curve. So, we get: 5x​5 - 10x3 + x + 2(2x - 3) + 6 = 05x5 - 10x3 + x + 4x - 6 + 6 = 0 5x5 - 10x3 + 5x = 05x(x4 - 2x2 + 1) = 0x4 - 2x2 + 1 = 0x4 - x2 - x2 + 1 = 0x2 (x2 - 1) - 1(x2 - 1) = 0(x2 - 1)2 = 0x = 1Plugging in the value of x = 1 in equation (i). We get, y = 2 - 3 = -1. Hence, the required point is (1, -1).   RegardsArun (askIITians forum expert)
```
one year ago
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