find the equation of the normal to the curve y=3x2-2x-1which is perpendicular to the line y=x-3Solution:y = 3x^2 - 2x - 1 (1) Its derivative is dy/dx = 6x - 2Let the equation of tangent line be y = mx + c1 = x + c1 (2)(Tangent line is parallel to y = x - 3,...
百度试题 结果1 题目【题目】Find the equation of the normal to the curve a the point with the given x-coordinate.y=1-x^2wherex=0 相关知识点: 试题来源: 解析 【解析】x=0 反馈 收藏
The equation of the normal to the curve3x2−y2=8which is parallel tox+3y=8isx−3y=8(b)x−3y+8=0(c)x+3y±8=0(d)x+3y=0 View Solution The normal to the curvex2=4ypassing (1,2) is(A)x+y=3(B)x−y=3(C)x+y=1(D)x−y=1 ...
题目1.Find the equation of the tangent to the curve y=x2,which is parallel to the x-axis2.Find the equation of the tangent to the curve y=x2-2x which is perpendicular to the line 2y=x-13.Find the equation of the normal to the curve y=3x2-2x-1 which...
find the equation of the normal to the curve at te point with the given x-coordinate y=1-x2(x的2次方) where x=0 这道题我切线求斜率求得是0 如果对 那法线斜率多少 扫码下载作业帮搜索答疑一搜即得 答案解析 查看更多优质解析 解答一 举报 答案是对的法线斜率不存在,倾角90度 解析看不懂?免费...
结果1 题目6)Find the equation of the normal to the curve y=(3x)/(2x-3)atlneoig in2x-3 相关知识点: 试题来源: 解析 at the origin so gradient normal is 1 and the origin is (0,0) So the equation of the normal is 反馈 收藏 ...
【题目】如果切线的斜率为0那法线呢find the equation of the normal to the curve at te point with the given x-coordinate y=1-r2的2次方)where x=0这道题我切线求斜率求得是0如果对那法线斜率多少 相关知识点: 试题来源: 解析 【解析】答案是对的法线斜率不存在,倾角90度 结果一 题目 如果切线的...
Find an equation of the normal line to the curve {eq}y = \sqrt x {/eq} that is parallel to the line {eq}2x+y=1 {/eq}. Finding the Normal Line to a Curve: The slope of a normal and slope of a tangent line are interrelated to each other. ...
If the equation of the normal to the curve y=f(x) at x=0 is 3x-y+3=0 t... 04:03 If lim(xrarr1)((asin(x-1)+bcos(x-1)+4))/(x^2-1)=2, then (a,b) is equal... 02:32 If a gt 0 and lim(xrarr oo) {sqrt(x^2+x+1)-(ax+b)}=0, then (a,b) lies ......
Let m is the slope of tangent of the curve and n is the slope of normal to the curve. Then, m*n = -1=> n = -1/m. Now, equation of the curve, x^2+2y^2-4x-6y+8 = 0 Differentiating it w.r.t. x, =>2x+4ydy/dx - 4 -6dy/dx = 0 =>dy/dx = (4-2x)/(4y-6)...